Mc D Chapter 9 Solutions1

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529Algebra 2

Worked-Out Solution Key

Prerequisite Skills (p. 612)

1. The graph of a quadratic function is a parabola.

2. The graph of the rational function y 5 2 } x is a hyperbola.

3. Two equations of the form Ax 1 By 5 C and Dx 1 Ey 5 F form a linear system of equations.

4. y 5 x2 2 3 5. y 5 20.25x2

1

x

y

21

(0, 23)

x 5 0

1

x

y

22

x 5 0

(0, 0)

6. y 5 3(x 1 1)2 7. y 5 0.5(x 2 2)2 1 4

4

x

y

23 (21, 0)

x 5 21

1

x

y

21

(2, 4)

x 5 2

8. x2 2 4x 1 7 5 0 9. x2 2 8x 2 15 5 0

x2 – 4x 5 27 x2 2 8x 5 15

x2 2 4x 1 4 5 27 1 4 x2 2 8x 1 16 5 15 1 16

(x 2 2)2 5 23 (x 2 4)2 5 31

x 2 2 5 6 Ï}

23 x 2 4 5 6 Ï}

31

x 5 2 6 Ï}

23 x 5 4 6 Ï}

31

x 5 2 6 i Ï}

3

10. 3x2 1 9x 2 12 5 0

3x2 1 9x 5 12

3 1 x2 1 3x 1 9 } 4 2 5 12 1 3 1 9 }

4 2

3 1 x 1 3 } 2 2 2 5 12 1

27 } 4

3 1 x 1 3 } 2 2 2 5

75 } 4

1 x 1 3 } 2 2 2 5

25 } 4

x 1 3 } 2 5 6

5 } 2

x 5 2 3 } 2 6

5 } 2

x 5 24 or x 5 1

11. 2x 2 y 5 11 12. x 1 5y 5 217

2x 2 2y 5 23 22x 2 3y 5 13

x 2 3y 5 8 2x 1 2y 5 24

x 5 8 1 3y 2x 5 24 2 2y

2(8 1 3y) 2 y 5 11 x 5 4 1 2y

16 1 6y 2 y 5 11 (4 1 2y) 1 5y 5 217

5y 5 25 4 1 7y 5 217

y 5 21 7y 5 221

x 5 8 1 3(21) y 5 23

x 5 5 x 5 4 1 2(23)

The solution is (5, 21). x 5 22

The solution is (22, 23).

13. 24x 1 7y 5 214

2x 2 6y 5 12

22x 1 y 5 22

22x 5 22 2 y

x 5 1 1 y } 2

2 1 1 1 y } 2 2 2 6y 5 12

2 1 y 2 6y 5 12

25y 5 10

y 5 22

x 5 1 1 (22)

} 2

x 5 1 2 1

x 5 0


Lesson 9.1

9.1 Guided Practice (pp. 615–616)

1. d 5 Ï}}}

(21 2 3)2 1 (5 2 (23))2

5 Ï}

16 1 64

5 Ï}

80

5 4 Ï}

5

2. RS 5 Ï}}

(5 2 (21))2 1 (2 2 3)2 5 Ï}

36 1 1 5 Ï}

37

RT 5 Ï}}

(3 2 (21))2 1 (6 2 3)2 5 Ï}

16 1 9 5 Ï}

25 5 5

ST 5 Ï}}

(3 2 5)2 1 (6 2 2)2 5 Ï}

4 1 16 5 Ï}

20 5 2 Ï}

5

Because RS Þ RT Þ ST, nRST is scalene.

3. a. (0, 0), (24, 12)

M 1 0 1 (24) }

2 ,

0 1 12 }

2 2 5 M(22, 6)

The midpoint is (22, 6).

Chapter 9

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530Algebra 2Worked-Out Solution Key

b. m 5 12 2 0

} 24 2 0 5

12 }

24 5 23

Slope of perpendicular bisector:

2 1 } m 5 2

1 }

(23) 5

1 } 3

Point slope form:

y 2 6 5 1 } 3 (x 2 (22))

y 5 1 } 3 x 1

2 } 3 1 6

y 5 1 } 3 x 1

20 } 3

4. a. (22, 1), (4, 27)

M 1 22 1 4 }

2 ,

1 1 (27) }

2 2 5 M (1, 23)


b. m 5 27 2 1

} 4 2 (22)

5 28

} 6 5 2 4 } 3


2 1 } m 5 2

1 }

1 2 4 } 3 2 5

3 } 4

Point slope form:

y 2 (23) 5 3

} 4 (x 2 1)

y 1 3 5 3 } 4 x 2

3 } 4

y 5 3 } 4 x 2

15 } 4

5. a. (3, 8), (25, 210)

M 1 3 1 (25) }

2 ,

8 1 (210) }

2 2 5 M 1 22

} 2 ,

22 }

2 2 5 M (21, 21)


b. m 5 210 2 8

} 25 2 3 5

218 }

28 5 9 } 4


2 1 } m 5 2

1 }

9 } 4 5 2

4 } 9

Point slope form:

y 2 (21) 5 2 4 } 9 (x 2 (21))

y 1 1 5 2 4 } 9 x 2

4 } 9

y 5 2 4 } 9 x 2

13 } 9

6.

x

y

4

20O (0, 0)

B (16, 8)

A (6, �2)

C

Let O represent (0, 0), A represent (6, 22), B represent (16, 8), and C represent the center of the circle.

Midpoint of }

OA :

M 1 6 1 0 }

2 ,

22 1 0 }

2 2 5 M (3, 21)

Slope of }

OA :

M 5 22 2 0

} 6 2 0 5 2 1 } 3


2 1 } m 5 2

1 }

1 2 1 } 3 2

5 3

Perpendicular bisector of }

OA :

y 2 (21) 5 3(x 2 3)

y 1 1 5 3x 2 9

y 5 3x 2 10

Midpoint of }

AB :

M 1 6 1 16 }

2 ,

22 1 8 }

2 2 5 M (11, 3)

Slope of }

AB :

m 5 8 2 (22)

} 16 2 6 5 10

} 10 5 1


2 1 } m 5 2

1 } 1 5 21


AB :

y 2 3 5 21(x 2 11)

y 2 3 5 2x 1 11

y 5 2x 1 14

Intersection of two perpendicular bisectors:

3x 2 10 5 2x 1 14

4x 5 24

x 5 6

y 5 26 1 14

y 5 8

The center of the circle is C (6, 8).

OC 5 Ï}}

(6 2 0)2 1 (8 2 0)2 5 Ï}

36 1 64 5 10

The diameter is 2(10) 5 20.

9.1 Exercises (pp. 617–619)

Skill Practice

1. The distance d between (x1, y1) and (x2, y2) is

d 5 Ï}}

(x2 2 x1)2 1 ( y2 2 y1)2 .

The midpoint of the line segment joining (x1, y1),

and (x2, y2) is M 1 x1 1 x2 }

2 ,

y1 1 y2 }

2 2 .

2. When fi nding the midpoint of a line segment joining two points, it does not matter which point you choose as (x1, y1), because addition is a commutative operation.

Chapter 9, continued

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531Algebra 2


3. (0, 0), (8, 15)

d 5 Ï}}

(8 2 0)2 1 (15 2 0)2 5 Ï}

64 1 225 5 Ï}

289 5 17

M 1 0 1 8 } 2 ,

0 1 15 }

2 2 5 1 4,

15 }

2 2

4. (0, 0), (4, 2)

d 5 Ï}}

(4 2 0)2 1 (2 2 0)2 5 Ï}

16 1 4 5 Ï}

20 5 2 Ï}

5

M 1 0 1 4 } 2 ,

0 1 2 }

2 2 5 (2, 1)

5. (0, 6), (5, 24)

d 5 Ï}}

(5 2 0)2 1 (24 2 6)2

5 Ï}

25 1 100

5 Ï}

125

5 5 Ï}

5

M 1 0 1 5 }

2 ,

6 1 (24) }

2 2 5 1 5 } 2 , 1 2

6. (27, 0), (5, 3)

d 5 Ï}}

(5 2 (27))2 1 (3 2 0)2

5 Ï}

(12)2 1 (3)2

5 Ï}

144 1 9

5 Ï}

153

5 3 Ï}

17

M 1 27 1 5 }

2 ,

0 1 3 }

2 2 5 1 21,

3 }

2 2

7. (2,21), (6,25)

d 5 Ï}}

(6 2 2)2 1 (25(21))2

5 Ï}

42 1 (24)2

5 Ï}

16 1 16

5 Ï}

32

5 4 Ï}

2

M 1 2 1 6 }

2 ,

21 1 (25) }

2 2 5 1 8 }

2 ,

26 }

2 2 5 (4, 23)

8. (21, 22), (8, 4)

d 5 Ï}}}

(8 2 (21))2 1 (4 2 (22))2

5 Ï}

92 1 62

5 Ï}

81 1 36

5 Ï}

117

5 3 Ï}

13

M 1 21 1 8 }

2 ,

22 1 4 }

2 2 5 1 7 } 2 , 1 2

9. (24, 8), (8, 24)

d 5 Ï}}}

(8 2 (24))2 1 (24 2 8)2

5 Ï}}

122 1 (212)2

5 Ï}

144 1 144

5 Ï}

288

5 12 Ï}

2

M 1 24 1 8 }

2 ,

8 1 (24) }

2 2 5 (2, 2)

10. (6, 23), (10, 29)

d 5 Ï}}}

(10 2 6)2 1 (29 2 (23))2

5 Ï}

42 1 (26)2

5 Ï}

16 1 36

5 Ï}

52

5 2 Ï}

13

M 1 6 1 10 }

2 ,

23 1 (29) }

2 2 5 (8, 26)

11. (24, 4), (5, 24)

d 5 Ï}}}

(5 2 (24))2 1 (24 2 4)2

5 Ï}

92 1 (28)2

5 Ï}

81 1 64

5 Ï}

145

M 1 24 1 5 } 2 ,

4 1 (24) }

2 2 5 1 1 } 2 , 0 2

12. (11, 212), (18, 12)

d 5 Ï}}}

(18 2 11)2 1 (12 2 (212))2

5 Ï}

72 1 242

5 Ï}

49 1 576

5 Ï}

625

5 25

M 1 11 1 18 } 2 ,

212 1 12 }

2 2 5 1 29

} 2 , 0 2 13. (25, 1), (15, 8)

d 5 Ï}}}

(15 2 (25))2 1 (8 2 1)2

5 Ï}

202 1 72

5 Ï}

400 1 49

5 Ï}

449

M 1 25 1 15 } 2 ,

1 1 8 }

2 2 5 1 5,

9 }

2 2

14. (9, 9), (216, 216)

d 5 Ï}}}

(216 2 9)2 1 (216 2 9)2

5 Ï}}

(225)2 1 (225)2

5 Ï}

625 1 625

5 Ï}

1250

5 25 Ï}

2

M 1 9 1 (216) } 2 ,

9 1 (216) }

2 2 5 1 27

} 2 , 27

} 2 2 5 1 2

7 } 2 , 2

7 } 2 2

15. (23.8, 15), (6.2, 211)

d 5 Ï}}}

(6.2 2 (23.8))2 1 (211 2 15)2

5 Ï}}

102 1 (226)2

5 Ï}

100 1 676

5 Ï}

776

5 2 Ï}

194

M 1 23.8 1 6.2 } 2 ,

15 1 (211) }

2 2 5 1 2.4

} 2 , 4 }

2 2 5 (1.2, 2)


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16. (1.5, 4), (2.3, 9)

d 5 Ï}}

(2.3 2 1.5)2 1 (9 2 4)2

5 Ï}

0.82 1 52

5 Ï}

0.64 1 25

5 Ï}

641

} 25

5 Ï}

641 } 5

ø 5.06

M 1 1.5 1 2.3 } 2 ,

4 1 9 }

2 2 5 1 3.8

} 2 , 13

} 2 2 5 (1.9, 6.5)

17. (22.4, 26.7), (3.1, 25.3)

d 5 Ï}}}}

(3.1 2 (22.4))2 1 (25.3 2 (26.7))2

5 Ï}

5.52 1 1.42

5 Ï}}

30.25 1 1.96

5 Ï}

32.21

5 Ï}

3221 } 10

ø 5.68

M 1 22.4 1 3.1 } 2 ,

26.7 1 (25.3) }}

2 2 5 1 0.7

} 2 , 212

} 2 2 5 (0.35, 26)

18. C; (24, 3), (6, 6)

d 5 Ï}}

(6 2 (24))2 1 (6 2 3)2 5 Ï}

102 1 32 5 Ï}

109

19. A; (23, 7), (5, 22)

M 1 5 1 (23) } 2 ,

7 1 (22) }

2 2 5 1 2 } 2 ,

5 }

2 2 5 1 1,

5 }

2 2

20. The difference between the y-coordinates should be (6 2 (21)), not (6 2 1).

d 5 Ï}}

(2 2 5)2 1 (6 2 (21))2 5 Ï}

9 1 49 5 Ï}

58

21. In the distance formula, the squares of the distances should be added, not subtracted.

d 5 Ï}}

(2 2 (24))2 1 (8 2 3)2 5 Ï}

36 1 25 5 Ï}

61

22. A(25, 0), B(0, 6), C(5, 0)

AB 5 Ï}}

(0 2 (25))2 1 (6 2 0)2 5 Ï}

25 1 36 5 Ï}

61

AC 5 Ï}}

(5 2 (25))2 1 (0 2 0)2 5 Ï}

102 5 10

BC 5 Ï}}

(5 2 0)2 1 (0 2 6)2 5 Ï}

52 1 (26)2

5 Ï}

25 1 36 5 Ï}

61

Because AB 5 BC, nABC is isosceles.

23. A(0, 23), B(0, 3), C(3, 0)

AB 5 Ï}}

(0 2 0)2 1 (3 2 (23))2 5 Ï}

0 1 62 5 6

AC 5 Ï}}

(3 2 0)2 1 (0 2 (23))2 5 Ï}

32 1 32

5 Ï}

18 5 3 Ï}

2

BC 5 Ï}}

(3 2 0)2 1 (0 2 3)2 5 Ï}

32 1 (23)2

5 Ï}

18 5 3 Ï}

2

Because AC 5 BC, nABC is isosceles.

24. A(3, 5), B(5, 23), C(7, 23)

AB 5 Ï}}

(5 2 3)2 1 (23 2 5)2 5 Ï}

22 1 (28)2

5 Ï}

4 1 64 5 Ï}

68 5 2 Ï}

17

AC 5 Ï}}

(7 2 3)2 1 (23 2 5)2 5 Ï}

42 1 (28)2

5 Ï}

16 1 64 5 Ï}

80 5 4 Ï}

5

BC 5 Ï}}}

(7 2 5)2 1 (23 2 (23))2 5 Ï}

22 1 02

5 Ï}

4 5 2

Because AB Þ AC Þ BC, nABC is scalene.

25. A(22, 5), B(1, 21), C(4, 6)

AB 5 Ï}}}

(1 2 (22)2 1 (21 2 5)2

5 Ï}

32 1 (26)2

5 Ï}

9 1 36 5 Ï}

45 5 3 Ï}

5

AC 5 Ï}}

(4 2 (22))2 1 (6 2 5)2

5 Ï}

62 1 12

5 Ï}

36 1 1 5 Ï}

37

BC 5 Ï}}

(4 2 1)2 1 (6 2 (21)2

5 Ï}

32 1 72

5 Ï}

9 1 49 5 Ï}

58


26. A(1, 4), B(4, 1), C(7, 4)

AB 5 Ï}}

(4 2 1)2 1 (1 2 4)2

5 Ï}

32 1 (23)2

5 Ï}

9 1 9 5 Ï}

18 5 3 Ï}

2

AC 5 Ï}}

(7 2 1)2 1 (4 2 4)2

5 Ï}

62 5 6

BC 5 Ï}}

(7 2 4)2 1 (4 2 1)2

= Ï}

32 1 32

5 Ï}

9 1 9 5 Ï}

18 5 3 Ï}

2


27. A(24, 1), B(22, 6), C(0, 21)

AB 5 Ï}}}

(22 2 (24))2 1 (6 2 1)2

5 Ï}

22 1 52

5 Ï}

4 1 25 5 Ï}

29

AC 5 Ï}}}

(0 2 (24))2 1 (21 2 1)2

5 Ï}

42 1 (22)2

5 Ï}

16 1 4

5 Ï}

20 5 2 Ï}

5

BC 5 Ï}}}

(0 2 (22))2 1 (21 2 6)2

5 Ï}

22 1 (27)2

5 Ï}

4 1 49 5 Ï}

53



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533Algebra 2


28. A(21, 26), B(1, 1), C(4, 25)

AB 5 Ï}}}

(1 2 (21))2 1 (1 2 (26))2

5 Ï}

22 1 72 5 Ï}

4 1 49 5 Ï}

53

AC 5 Ï}}}

(4 2 (21))2 1 (25 2 (26))2

5 Ï}

52 1 12 5 Ï}

25 1 1 5 Ï}

26

BC 5 Ï}}

(4 2 1)2 1 (25 2 1)2

5 Ï}

32 1 (26)2

5 Ï}

9 1 36

5 Ï}

45 5 3 Ï}

5


29. A(24, 3), B(2, 21), C(8, 21)

AB 5 Ï}}}

(2 2 (24))2 1 (21 2 3)2

5 Ï}

62 1 (24)2

5 Ï}

36 1 16

5 Ï}

52 5 2 Ï}

13

AC 5 Ï}}}

(8 2 (24))2 1 (21 2 3)2

5 Ï}}

122 1 (24)2

5 Ï}

144 1 16

5 Ï}

160 5 4 Ï}

10

BC 5 Ï}}}

(8 2 2)2 1 (21 2 (21))2 5 Ï}

62 1 0 5 6


30. A(3, 5), B(6, 9), C(11, 9)

AB 5 Ï}}

(6 2 3)2 1 (9 2 5)2

5 Ï}

32 1 42

5 Ï}

9 1 16

5 Ï}

25 5 5

AC 5 Ï}}

(11 2 3)2 1 (9 2 5)2

5 Ï}

82 1 42

5 Ï}

64 1 16

5 Ï}

80 5 4 Ï}

5

BC 5 Ï}}

(11 2 6)2 1 (9 2 9)2 5 Ï}

52 5 5


31. (3, 8), (7, 14)

M 1 3 1 7 } 2 ,

14 1 8 }

2 2 5 (5, 11)

m 5 14 2 8

} 7 2 3 5 6 } 4 5

3 } 2

2 1 } m 5 2

1 }

3 }

2 5 2

2 } 3

y 2 11 5 2 2 } 3 (x 2 5)

y 2 11 5 2 2 } 3 x 1

10 } 3

y 5 2 2 } 3 x 1

43 } 3

32. (25, 6), (1, 8)

M 1 25 1 1 } 2 ,

6 1 8 }

2 2 5 1 24

} 2 , 14

} 2 2 5 (22, 7)

m 5 8 2 6

} 1 2 (25)

5 2 } 6 5

1 } 3

2 1 } m 5 2

1 }

1 }

3 5 23

y 2 7 5 23(x 2 (22))

y 2 7 5 23x 2 6

y 5 23x 1 1

33. (23, 26), (21, 2)

M 1 23 1 (21) } 2 ,

26 1 2 }

2 2 5 1 24

} 2 , 24

} 2 2 5 (22, 22)

m 5 2 2 (26)

} 21 2 (23)

5 8 } 2 5 4

2 1 } m 5 2

1 } 4

y 2 (22) 5 2 1 } 4 (x 2 (22))

y 1 2 5 2 1 } 4 x 2

1 } 2

y 5 2 1 } 4 x 2

5 } 2

34. (1, 4), (6, 26)

M 1 1 1 6 } 2 ,

4 1 (26) }

2 2 5 1 7 } 2 , 21 2

m 5 26 2 4

} 6 2 1 5 210

} 5 5 22

2 1 } m 5 2

1 }

22 5 1 } 2

y 2 (21) 5 1 } 2 1 x 2

7 } 2 2

y 1 1 5 1 } 2 x 2

7 } 4

y 5 1 } 2 x 2

11 } 4

35. (23, 25), (9, 22)

M 1 23 1 9 } 2 ,

25 1 (22) }

2 2 5 1 6 } 2 ,

27 }

2 2 5 1 3, 2

7 } 2 2

m 5 22 2 (25)

} 9 2 (23)

5 3 } 12 5

1 } 4

2 1 } m 5 2

1 }

1 } 4 5 24

y 2 1 2 7 } 2 2 5 24(x 2 3)

y 1 7 } 2 5 24x 1 12

y 5 24x 1 17

} 2


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36. (5, 10), (10, 7)

M 1 5 1 10 } 2 ,

10 1 7 }

2 2 5 1 15

} 2 , 17

} 2 2

m 5 7 2 10

} 10 2 5 5 2 3 } 5

2 1 } m 5 2

1 }

2 3 } 5 5

5 } 3

y 2 17

} 2 5 5 } 3 1 x 2

15 } 2 2

y 2 17

} 2 5 5 } 3 x 2

75 } 6

y 5 5 } 3 x 2

75 } 6 1

51 } 6

y 5 5 } 3 x 2

24 } 6

y 5 5 } 3 x 2 4

37. 1 x1 1 x2 }

2 ,

y1 1 y2 }

2 2 5 (4, 2)

x1 1 x2

} 2 5 4 and

y1 1 y2 }

2 5 2

x1 1 x2 5 8 and y1 1 y2 5 4

Sample answer: (x1 , y1) 5 (6, 4) and (x2 , y2) 5 (2, 0)

38. A(8, 4), B(0, 0), C(10, 0)

Midpoint of }

BC : M 1 10 1 0 } 2 ,

0 1 0 }

2 2 5 (5, 0)

Slope of }

AM : m 5 0 2 4

} 5 2 8 5 24

} 23 5

4 } 3

Equation of median:

y 2 0 5 4 } 3 (x 2 5)

y 5 4 } 3 x 2

20 } 3

39. A(3, 10), B(4, 2), C(10, 8)

Midpoint of }

BC : M 1 4 1 10 } 2 ,

2 1 8 }

2 2 5 (7, 5)

Slope of }

AM : m 5 5 2 10

} 7 2 3 5 25

} 4 5 2 5 } 4

Equation of median:

y 2 5 5 2 5 } 4 (x 2 7)

y 2 5 5 2 5 } 4 x 1

35 } 4

y 5 2 5 } 4 x 1

55 } 4

40. A(2, 6), B(3, 1), C(7, 5)

Midpoint of }

BC : M 1 3 1 7 } 2 ,

1 1 5 }

2 2 5 (5, 3)

Slope of }

AM : m 5 3 2 6

} 5 2 2 5 23

} 3 5 21

Equation of median:

y 2 3 5 21(x 2 5)

y 2 3 5 2x 1 5

y 5 2x 1 8

41. (0, 3), (x, 5); d 5 2 Ï}

10

d 5 Ï}}

(x 2 0)2 1 (5 2 3)2

2 Ï}

10 5 Ï}

x2 1 22

Ï}

40 5 Ï}

x2 1 4

40 5 x2 1 4

36 5 x2

66 5 x

42. (23, 21), (2, y); d 5 Ï}

41

d 5 Ï}}}

(2 2 (23))2 1 (y 2 (21))2

Ï}

41 5 Ï}}

52 1 ( y 1 1)2

41 5 25 1 ( y 1 1)2

16 5 ( y 1 1)2

6 4 5 y 1 1

21 6 4 5 y

y 5 25 or 3

43. (x, 7), (24, 1); d 5 6 Ï}

2

d 5 Ï}}

(24 2 x)2 1 (1 2 7)2

6 Ï}

2 5 Ï}}

(24 2 x)2 1 (26)2

Ï}

72 5 Ï}}

(24 2 x)2 1 36

72 5 (24 2 x)2 1 36

36 5 (24 2 x)2

66 5 24 2 x

x 5 24 6 6

x 5 210 or 2

44. (1, y), (8, 13); d 5 Ï}

74

d 5 Ï}}

(8 2 1)2 1 (13 2 y)2

Ï}

74 5 Ï}}

72 1 (13 2 y)2

Ï}

74 5 Ï}}

49 1 (13 2 y)2

74 5 49 1 (13 2 y)2

25 5 (13 2 y)2

65 5 13 2 y

y 5 13 6 5

y 5 8 or 18


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535Algebra 2


45. (2, 3), (x, y); y 5 2x

d 5 Ï}}

(x 2 2)2 1 ( y 2 3)2

5 Ï}}

(x 2 2)2 1 (2x 2 3)2

5 Ï}}}

x2 2 4x 1 4 1 4x2 2 12x 1 9

5 Ï}}

5x2 2 16x 1 13

An equation for the distance is: d 5 Ï}}

5x2 2 16x 1 13 .

When d 5 Ï}

10 :

Ï}}

5x2 2 16x 1 13 5 Ï}

10

5x 2 16x 1 13 5 10

5x 2 16x 1 3 5 0

x 5 2(216) 6 Ï

}}

(216)2 2 4(5)(3) }}}

2(5)

x 5 16 6 Ï

}

256 2 60 }} 10

x 5 16614

} 10 5 1 } 5 or 3

When x 5 1 } 5 : y 5 2 1 1 } 5 2 5

2 } 5

When x 5 3: y 5 2(3) 5 6

Two points on the line y 5 2x that are each Ï}

10 units

from (2, 3) are 1 1 } 5 , 2 } 5 2 and (3, 6).

46. Distance from M to (x1 , y1):

d1 5 Î}}}

1 x1 2 x1 1 x2

} 2 2 2 1 1 y1 2 y1 1 y2

} 2 2 2

5 Î}}

1 x1 }

2 2

x2 } 2 2

2 1 1 y1

} 2 2 y2

} 2 2 2

5 Î}}}

x1

2

} 4 2

x1x2 } 2 1

x22

} 4 1 y1

2

} 4 2 y1y2

} 2 1 y2

2

} 4

Distance from M to (x2 , y2):

d2 5 Î}}}

1 x2 2 x1 1 x2

} 2 2 2 1 1 y2 2 y1 1 y2

} 2 2 2

5 Î}}

1 x2 }

2 2

x1 } 2 2

2 1 1 y2

} 2 2 y2

} 2 2 2

5 Î}}}

x2

2

} 4 2

x1x2 } 2 1

x12

} 4 1 y2

2

} 4 2 y1y2

} 2 1 y1

2

} 4

d1 5 d2

Equation of line containing (x1 , y1) and (x2 , y2):

y 2 y1 5 1 y2 2 y1

} x2 2 x1 2 (x 2 x1 )

y 5 1 y2 2 y1

} x2 2 x1 2 (x 2 x1 ) 1 y1

When x 5 x1 1 x2

} 2 :

y 5 1 y2 2 y1

} x2 2 x1 2 1

x1 1 x2 }

2 2 x1 2 1 y1

5 1 y2 2 y1

} x2 2 x1 2 1

x2 2 x1 }

2 2 1 y1

5 y2 2 y1

} 2 1 y1

5 y2 2 y1 1 2y1

}} 2

5 y1 1 y2

} 2

Because the point M satisfi es the equation, it lies on the line containing (x1, y1 ) and (x2, y2 ). Because d1 5 d2, M is equidistant from (x1, y1) and (x2, y2). So M is the midpoint of the line segment with endpoints (x1, y1) and (x2, y2).

Problem Solving

47. (6, 11), (22, 26)

d 5 Ï}}}

(22 2 6)2 1 (26 2 11)2

5 Ï}}

(28)2 1 (15)2

5 Ï}

64 1 225

5 Ï}

289

5 17

The robot moves 17 meters.

48. Memorial Medical: (0, 0.5)

Capital Airport: (21.1, 2.8)

d 5 Ï}}}

(0 2 (21.1))2 1 (0.5 2 2.8)2

5 Ï}}

1.12 1 (22.3)2

5 Ï}

1.21 1 5.29

5 Ï}

6.5

ø 2.55

From the Capital Airport, the medevac would have to fl y about 2.55 miles.


University of Illinois: (2.2, 25.1)

d 5 Ï}}}

(0 2 2.2)2 1 (0.5 2 (25.1))2

5 Ï}}

(22.2)2 1 5.62

5 Ï}}

4.84 1 31.36

5 Ï}

36.2 ø 6.02

From the University of Illinois, the medevac would have to fl y about 6.02 miles.


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Washington Park: (21.4, 20.8)

d 5 Ï}}}

(0 2 (21.4))2 1 (0.5 2 (20.8))2

5 Ï}

1.42 1 1.32

5 Ï}

1.96 1 1.69

5 Ï}

3.65 ø 1.91

From Washington Park, the medevac would have to fl y about 1.91 miles.


Spaulding Dam: (2.9, 23)

d 5 Ï}}}

(0 2 2.9)2 1 (0.5 2 (23))2

5 Ï}}

(22.9)2 1 3.52

5 Ï}}

8.41 1 12.25

5 Ï}

20.66

ø 4.55

From Spaulding Dam, the medevac would have to fl y about 4.55 miles.

52.

x

y3

3Home(0, 0)

(5, −2)Office

5

−42

d 5 Ï}}

(0 2 5)2 1 (0 2 (22))2

5 Ï}

(25)2 1 22

5 Ï}

25 1 4

5 Ï}

29

ø 5.39

It is about 5.39 miles between Li’s home and her offi ce.

53. a. M 1 3 1 (26) } 2 ,

11 1 5 }

2 2 5 1 2

3 } 2 , 8 2

b. Distance traveled from V to M 5 VS 1 SM.

VS 5 Ï}}

(26 2 0)2 1 (5 2 0)2 5 Ï}

36 1 25

5 Ï}

61 ø 7.81

SM 5 Î}}}

1 2 3 } 2 2 (26) 2 2 1 (8 2 5)2

5 Î} 1 9 } 2 2 2 1 32

5 Î} 81

} 4 1 9

5 Î}

117

} 4

5 Ï}

117 } 2 ø 5.41

VS 1 SM ø 7.81 1 5.41 5 13.22

The total distance traveled from V to M is about 13.22(0.1) ø 1.32 miles.

c. Distance from M to V through P 5 MP 1 PV.

MP 5 Î}}}

1 3 2 1 2 3 } 2 2 2 2 1 (11 2 8)2

5 Î} 1 9 } 2 2 2 1 32

5 Î} 81

} 4 1 9

5 Î}

117

} 4

5 Ï}

117 } 2 ø 5.41

PV 5 Ï}}

(3 2 0)2 1 (11 2 0)2

5 Ï}

32 1 112

5 Ï}

9 1 121

5 Ï}

130

ø 11.4

MP 1 PV ø 5.41 1 11.4 5 16.81

The total distance from M to V through P is about 16.81(0.1) ø 1.68 miles.

54. A(24, 2), O(0, 0), B(6, 4)

x

y

2

2

A(−4, 2)

O (0, 0)

B (6, 4)

Slope of }

AO :

m 5 2 2 0

} 24 2 0 5 2

1 } 2

2 1 } m 5 2

1 }

2 1 } 2 5 2

Midpoint of }

AO : M 1 24 1 0 } 2 ,

2 1 0 }

2 2 5 (22, 1)


AO :

y 2 1 5 2(x 2 (22))

y 2 1 5 2x 1 4

y 5 2x 1 5

Slope of }

BO :

m 5 4 2 0

} 6 2 0 5 2 } 3

2 1 } m 5 2

1 }

2 }

3 5 2

3 } 2

Midpoint of }

BO : M 1 6 1 0 } 2 ,

4 1 0 }

2 2 5 (3, 2)


BO :

y 2 2 5 2 3 } 2 (x 2 3)

y 2 2 5 2 3 } 2 x 1

9 } 2

y 5 2 3 } 2 x 1

13 } 2


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537Algebra 2


Intersection of perpendicular bisectors:

2x 1 5 5 2 3 } 2 x 1

13 } 2

7 }

2 x 5

3 } 2

x 5 3 } 7

y 5 2 1 3 } 7 2 1 5 5 41

} 7

The center of the circle is 1 3 } 7 , 41

} 7 2 . Distance from center to (0, 0):

d 5 Î}}

1 3 } 7 2 0 2 2 1 1 41 } 7 2 0 2 2

5 Î}}

1 3 } 7 2 2 1 1 41 } 7 2 2

5 Î} 9 }

49 1

1681 } 49

5 Î} 1690

} 49

5 13 Ï

}

10 } 7

ø 5.873

The original diameter of the dish was about 5.873(2) ø 11.75 inches.

55. A(2220, 220), O(0, 0), B(200, 40)

x

y

100

50

A(−220, 220)

B(200, 40)

O (0, 0)

Slope of }

AO :

m 5 220 2 0

} 2220 2 0 5 21

2 1 } m 5 2

1 }

21 5 1

Midpoint of }

AO : M 1 2220 1 0 }

2 ,

220 1 0 }

2 2 5 (2110, 110)


AO :

y 2 110 5 1(x 2 (2110))

y 2 110 5 x 1 110

y 5 x 1 220

Slope of }

BO :

m 5 40 2 0

} 200 2 0 5 1 } 5

2 1 } m 5 2

1 }

1 } 5 5 25

Midpoint of }

BO : M 1 200 1 0 } 2 ,

40 1 0 }

2 2 5 (100, 20)


BO :

y 2 20 5 25(x 2 100)

y 2 20 5 25x 1 500

y 5 25x 1 520


x 1 220 5 25x 1 520

6x 5 300

x 5 50

y 5 x 1 220

y 5 50 1 220 5 270

The center is (50, 270).

Distance from center to (0, 0):

d 5 Ï}}

(50 2 0)2 1 (270 2 0)2

5 Ï}}

2500 1 72,900

5 Ï}

75,400 ø 274.6

The diameter of the crater is about 2(274.6) 5 549.2 ø 550 feet.

56. a. M1 1 x1 1 0 } 2 ,

y1 1 0 }

2 2 5 1 x1

} 2 , y1

} 2 2

M2 1 x1 1 x2 } 2 ,

y1 1 0 }

2 2 5 1 x1 1 x2

} 2 , y1

} 2 2

b. Length of strip:

d 5 Ï}}}

1 x1 1 x2 }

2 2

x1 } 2 2

2

1 1 y1 } 2 2

y1 } 2 2

2

5 Î} 1 x2 }

2 2 2 1 02 5

x2 } 2

Length of sail’s base:

d 5 Ï}}

(x2 2 0)2 1 (0 2 0)2 5 Ï}

(x2)2 5 x2

The length of the strip is half the length of the base.

c. No; regardless of the values of x1, x2, and y1, the

length of the strip will always be x2

} 2 .

57. The car is in range of the tower when the distance d between the car and the tower is less than or equal to 50 miles.

Distance to tower from any point (x, 40):

d 5 Ï}}

(x 2 0)2 1 (40 2 0)2

d 5 Ï}

x2 1 1600

Horizontal distance traveled 5 60t

So, x 5 60t 2 100.

d 5 Ï}}

(60t 2 100)2 1 1600

d 5 Ï}}}

3600t 2 2 12,000t 1 10,000 1 1600

d 5 Ï}}}

3600t 2 2 12,000t 1 11,600

d 5 Ï}}

400(9t 2 2 30t 1 29)

d 5 20 Ï}}

9t 2 2 30t 1 29


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When d ≤ 50:

x

y100

25

(−100, 40)

(0, 0)

Car60 mph

Tower

N

SW E

d

20 Ï}}

9t 2 2 30t 1 29 ≤ 50

Ï}}

9t 2 2 30t 1 29 ≤ 5 } 2

9t 2 2 30t 1 29 ≤ 25 } 4

36t 2 2 120t 1 116 ≤ 25

36t 2 2 120t 1 91 ≤ 0 (6t 2 7)(6t 2 13) ≤ 0 6t 2 7 ≥ 0 and 6t 2 13 ≤ 0

t ≥ 7 } 6 and t ≤ 13 } 6

So, the car is in range of the tower when 7 }

6 ≤ t ≤ 13 }

6 , or

between about 1.17 hours and 2.17 hours.

Mixed Review

58. y 5 3x2 59. y 5 24x2

2

x

y

21

1

x

y

21

60. y 5 22x2 61. y 5 1 } 2 x2

1

x

y

21

1

x

y

21

62. y 5 2 1 } 3 x2 63. y 5

3 } 4 x2

1

x

y

22

1

x

y

21

64. x2 2 6x 2 16 5 (x 2 8)(x 1 2)

65. x2 2 15x 1 54 5 (x 2 9)(x 2 6)

66. 8x2 1 37x 1 20 5 (8x 1 5)(x 1 4)

67. 12x2 2 13x 1 3 5 (4x 2 3)(3x 2 1)

68. x3 1 5x2 2 9x 2 45 5 x2(x 1 5) 2 9(x 1 5)

5 (x 1 5)(x2 2 9) 5 (x 1 5)(x 1 3)(x 2 3)

69. x3 2 64 5 (x 2 4)(x2 1 4x 1 16) 70. Ï

}

3x 1 1 5 10 71. Ï}

5x 2 8 5 Ï}

x 1 12

3x 1 1 5 100 5x 2 8 5 x 1 12

3x 5 99 4x 5 20

x 5 33 x 5 5

72. Ï}

x 2 3 5 x 2 5

x 2 3 5 (x 2 5)2

x 2 3 5 x2 2 10x 1 25

0 5 x2 2 11x 1 28

0 5 (x 2 4)(x 2 7)

x 2 4 5 0 or x 2 7 5 0

x 5 4 or x 5 7

Check:

Ï}

4 2 3 0 4 2 5 Ï}

7 2 3 0 7 2 5

1 Þ 21 Ï}

4 0 2

2 5 2 ✓

The solution is 7.

73. x 2 3

} x 2 1

5 6 } x 14

(x 2 3)(x 1 4) 5 6(x 2 1)

x2 1 x 2 12 5 6x 2 6

x2 2 5x 2 6 5 0

(x 2 6)(x 1 1) 5 0

x 2 6 5 0 or x 1 1 5 0

x 5 6 or x 5 21

Check:

6 2 3

} 6 2 1

0 6 }

6 1 4

21 2 3 }

21 2 1 0

6 }

21 1 4

3 } 5 0

6 } 10

24 }

22 0

6 }

3

3 } 5 5

3 } 5 ✓ 2 5 2 ✓

The solutions are 6 and 21.

74. x 2 6

} x 2 3

5 x 2 2

} x 1 4

(x 2 6)(x 1 4) 5 (x 2 2)(x 2 3)

x2 2 2x 2 24 5 x2 2 5x 1 6

3x 5 30

x 5 10

75. 1 }

x 1 1 1 3 5

4x } x 1 1

(x 1 1) 1 1 }

x 1 1 1 3 2 5 4x

1 1 3(x 1 1) 5 4x

1 1 3x 1 3 5 4x

4 5 x


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539Algebra 2


Lesson 9.2

9.2 Guided Practice (pp. 622– 623)

1. y2 5 26x

( )32 , 02 1

x

y

21

x 5 32

Focus: ( p, 0)

26x 5 4px

26 5 4p

2 3 } 2 5 p

Focus: 1 2 3 } 2 , 0 2

Directrix: x 5 2p, or x 5 3 } 2

Axis of symmetry: horizontal ( y 5 0)

2. x2 5 2y

( )12

12

0,

y 5 2

1x

y

Focus: (0, p)

x2 5 4py

2y 5 4py

1 }

2 5 p

Focus: 1 0, 1 }

2 2

Directrix: y 5 2p, or y 5 2 1 } 2

Axis of symmetry: vertical (x 5 0)

3. y 5 2 1 } 4 x2

y 5 1

(0, 21)

2

x

y

22

x2 5 24y

Focus: (0, p)

24y 5 4py

21 5 p

Focus: (0, 21)

Directrix: y 5 2p, or y 5 1

Axis of symmetry: vertical (x 5 0)

4. x 5 1 } 3 y2

x

y

( )34

34

, 0

x 5 2

y2 5 3x

Focus: ( p, 0)

3x 5 4px

3 }

4 5 p

Focus: 1 3 } 4 , 0 2

Directrix: x 5 2p, or x 5 2 3 } 4

Axis of symmetry: horizontal ( y 5 0)

5. Directrix: y 5 2p 5 2

p 5 22

Equation of parabola: x2 5 4py

x2 5 4(22)y

x2 5 28y

6. Directrix: x 5 2p 5 4

p 5 24

Equation of parabola: y2 5 4px

y2 5 4(24) x

y2 5 216x

7. Focus: (22, 0)

p 5 22

Equation of parabola: y2 5 4px

y2 5 4(22) x

y2 5 28x

8. Focus: (0, 3)

p 5 3

Equation of parabola: x2 5 4py

x2 5 4(3)y

x2 5 12y

9. Vertex: (0, 0)

Focus: (10, 0)

p 5 10

The parabola opens right, so its equation has the form y2 5 4px.

y2 5 4px

y2 5 4(10)x

y2 5 40x

The antenna’s depth is the x-value at the outside edge.

Radius 5 1 } 2 (16) 5 8. The antenna extends 8 feet to either

side of the vertex (0, 0).

When y 5 8:

(8)2 5 40x

64 5 40x

1.6 5 x

The antenna is 1.6 feet deep.


Skill Practice

1. A parabola is the set of all points in a plane equidistant from a point called the focus and a line called the directrix.

2. Sample answer: x2 5 4py opens up or down (depending on the value of p) while y2 5 4px opens left or right (depending on the value of p).

3. y2 5 16x

(4, 0)1

1 x

y

x 5 24

4px 5 16x

p 5 4

Focus: (4, 0)

Directrix: x 5 2p 5 24

Axis of symmetry: y 5 0


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4. x2 5 26y y 5

32

0, 232( )

1

x

y

23

4py 5 26y

p 5 2 6 } 4 5 2

3 } 2

Focus: 1 0, 2 3 } 2 2

Directrix: y 5 2p 5 3 } 2

Axis of symmetry: x 5 0

5. x2 5 20y

2

2

y 5 25

(0, 5)

x

y

4py 5 20y

p 5 5

Focus: (0, 5)

Directrix: y 5 2p 5 25


6. y2 5 28x

x 5 27(7, 0)

222

x

y

4px 5 28x

p 5 7

Focus: (7, 0)



7. y2 5 210x

1

1 x

y

x 5 52

2 , 052( )

4px 5 210x

p 5 2 10

} 4 5 2 5 } 2

Focus: 1 2 5 } 2 , 0 2

Directrix: x 5 2p 5 5 } 2


8. x2 5 30y y

x4

4

y 5 2152

( )1520 ,

4 py 5 30y

p 5 30

} 4 5 15

} 2

Focus: 1 0, 15

} 2 2

Directrix: y 5 2p 5 2 15

} 2


9. y2 5 22x

x 5 12

( )12 , 02

2

x

y

21

4px 5 22x

p 5 2 1 } 2

Focus: 1 2 1 } 2 , 0 2



10. x2 5 236y

x

y

3

3

(0, 29)

y 5 9 4py 5 236y

p 5 2 36

} 4 5 29

Focus: (0, 9)



11. x2 5 12y

x

y

2

2

(0, 3)

y 5 23

4py 5 12y

p 5 3

Focus: (0, 3)


Axis of symmetry:x 5 0

12. 22y 5 x2

x

y1

1

y 5 12

(0, 2 )12

22y 5 4py

2 1 } 2 5 p

Focus: 1 0, 2 1 }

2 2



13. x 5 4y2

x

y

0.2

( )116

116

, 0

x 5 2

0.1 y2 5

1 } 4 x

4px 5 1 } 4 x

p 5 1 } 16

Focus: 1 1 } 16

, 0 2

Directrix: x 5 2p 5 2 1 } 16


14. 2x2 5 48y

x

y

5

5

y 5 12

(0, 212)

x2 5 248y

4py 5 248y

p 5 2 48

} 4 5 212

Focus: (0, 212)



15. 5x2 5 215y

x

y

0.5

1.5

y 5 34

34( ), 20

x2 5 23y

4py 5 23y

p 5 2 3 } 4

Focus: 1 0, 2 3 } 4 2




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541Algebra 2


16. 2y2 5 18x

x

y

2

2

x 5 92( )9

2 , 02 y2 5 218x

4px 5 218x

p 5 2 18

} 4 5 2 9 } 2

Focus: 1 2 9 } 2 , 0 2



17. 224x 5 3y2

1

x 5 2

(22, 0) 1

x

y

y2 5 28x

4px 5 28x

p 5 22

Focus: (22, 0)



18. 14x 5 6y2

1

x 5 2

( )

712

712

, 01

x

y

y2 5 14

} 6 x

4px 5 7 } 3 x

p 5 7 } 12

Focus: 1 7 } 12

, 0 2

Directrix: x 5 2p 5 2 7 } 12


19. 1 }

8 x2 2 y 5 0

y 5 22

(0, 2)

1

1

x

y

1 }

8 x2 5 y

x2 5 8y

4py 5 8y

p 5 2

Focus: (0, 2)



20. 4x 2 11y2 5 0

x

y

0.1x 5 2

111

( )111

, 00.1 211y2 5 24x

y2 5 4 } 11 x

4px 5 4 } 11 x

p 5 1 } 11

Focus: 1 1 } 11

, 0 2

Directrix: x 5 2p 5 2 1 } 11


21. 5x2 1 12y 5 0

1

x

y

2

y 5 35

( )35, 20

5x2 5 212y

x2 5 2 12

} 5 y

4py 5 2 12

} 5 y

p 5 2 3 } 5

Focus: 1 0, 2 3 } 5 2



22. 25x 1 1 } 3 y2 5 0

x 5 23 34

( )343 , 01

x

y

21

1 }

3 y2 5 5x

y2 5 15x

4px 5 15x

p 5 15

} 4 5 3 3 }

4

Focus: 1 3 3 }

4 , 0 2

Directrix: x 5 2p 5 23 3 }

4


23. The parabola should open to the right rather than up.

26x 1 y2 5 0

x

y

1

1

( )32

32

, 0

x 5 2

y2 5 6x

4px 5 6x

p 5 3 } 2

Focus: 1 3 } 2 , 0 2

Directrix: x 5 2p 5 2 3 } 2


24. The parabola should open to the left rather than to the right.

0.5y2 1 x 5 0

x

y

22

1

x 5 12

( )12

, 02

0.5y2 5 2 x

y2 5 22x

4px 5 22x

p 5 2 1 } 2

Focus: 1 2 1 } 2 , 0 2




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25. D; 15y 1 3x2 5 0

3x2 5 215y

x2 5 25y

4py 5 25y

p 5 2 5 } 4

Directrix: y 5 2p 5 5 } 4 5 1.25

26. Focus: (2, 0) 27. Focus: (25, 0)

p 5 2 p 5 25

y2 5 4px y2 5 4(25)x

y2 5 4(2)x y2 5 220x

y2 5 8x

28. Focus: (3, 0) 29. Focus: (0, 24)

p 5 3 p 5 24

y2 5 4px x2 5 4py

y2 5 4(3)x x2 5 4(24)y

y2 5 12x x2 5 216y

30. Focus: (0, 8) 31. Focus: (0, 210)

p 5 8 p 5 210

x2 5 4py x2 5 4py

x2 5 4(8)y x2 5 4(210)y

x2 5 32y x2 5 240y

32. Focus: (0, 26) 33. Focus: (29, 0)

p 5 26 p 5 29

x2 5 4py y2 5 4px

x2 5 4(26)y y2 5 4(29)x

x2 5 224y y2 5 236x

34. Focus: 1 0, 7 }

4 2 35. Focus: 1 0, 2

3 } 8 2

p 5 7 } 4 p 5 2

3 } 8

x2 5 4py x2 5 4py

x2 5 4 1 7 } 4 2 y x2 5 4 1 2

3 } 8 2 y

x2 5 7y x2 5 2 3 } 2 y

36. Focus: 1 5 } 2 , 0 2 37. Focus: 1 2

9 } 16 , 0 2

p 5 5 } 2 p 5 2

9 } 16

y2 5 4px y2 5 4px

y2 5 4 1 5 } 2 2 x y2 5 4 1 2

9 } 16 2 x

y2 5 10x y2 5 2 9 } 4 x

38. A; Focus: (28, 0) 39. Directrix: x 5 3

p 5 28 p 5 23

y2 5 4px y2 5 4px

y2 5 4(28)x y2 5 4(23)x

y2 5 232x y2 5 212x

40. Directrix: y 5 27 41. Directrix: x 5 25

p 5 7 p 5 5

x2 5 4py y2 5 4px

x2 5 4(7)y y2 5 4(5)x

x2 5 28y y2 5 20x

42. Directrix: y 5 12 43. Directrix: y 5 24

p 5 212 p 5 4

x2 5 4py x2 5 4py

x2 5 4(212)y x2 5 4(4)y

x2 5 248y x2 5 16y

44. Directrix: x 5 22 45. Directrix: y 5 6

p 5 2 p 5 26

y2 5 4px x2 5 4py

y2 5 4(2)x x2 5 4(26)y

y2 5 8x x2 5 224y

46. Directrix: x 5 11

p 5 211

y2 5 4px

y2 5 4(211)x

y2 5 244x

47. Directrix: x 5 2 3 } 2 48. Directrix: y 5

5 } 12

p 5 3 } 2 p 5 2

5 } 12

y2 5 4px x2 5 4py

y2 5 4 1 3 } 2 2 x x2 5 4 1 2

5 } 12 2 y

y2 5 6x x2 5 2 5 } 3 y

49. Directrix: y 5 2 11

} 6 50. Directrix: x 5 2 1 } 18

p 5 11

} 6 p 5 1 } 18

x2 5 4py y2 5 4px

x2 5 4 1 11 }

6 2 y y2 5 4 1 1 }

18 2 x

x2 5 22

} 3 y y2 5 2 } 9 x

51. a. x2 5 ay

1

x

y

21

x2 5 yx2 5 4y

4p 5 a

p 5 a } 4

As a changes from 1 to 4, the new focus will be

located at (0, 1) rather than 1 0, 1 }

4 2 . The new directrix

will be y 5 21 rather than y 5 2 1 } 4 . The parabola will

be wider.


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543Algebra 2


b. y2 5 ax y2 5 2 x

y2 5 6x

12

1

x

y

22

4p 5 a

p 5 a } 4

As a changes from 6 to2 1 } 2 , the new focus will be

located at 1 2 1 } 8 , 0 2 rather than 1 3 }

2 , 0 2 . The new directrix

will be x 5 1 } 8 rather that x 5 2

3 } 2 . The parabola will be

narrower and will open left rather than right.

52. The equation x2 5 4py can be rewritten y 5 1 } 4p x2. So, if

x2 5 4py and y 5 ax2 represent the same parabola, then

y 5 1 }

4p x2 5 ax2 and a 5

1 } 4p .

53. As p increases, the focus of the parabola and the directrix move further away from the vertex, and the graph becomes wider. Because each point on a parabola is equidistant from the focus and the directrix, this has the effect of making the parabola wider and wider as p increases.

54. Focus: (0, p)

Directrix: y 5 2p

Distance from (x, y) to (0, p):

d 5 Ï}}

(x 2 0)2 1 ( y 2 p)2 5 Ï}}

x2 1 ( y 2 p)2

Distance from (x, y) to directrix y 5 2p:

d 5 Ï}}

(x 2 x)2 1 ( y 2 (2p))2 5 Ï}

( y 1 p)2

The point (x, y) is equidistant from the focus and directrix, so the above distances are equal.

x2 1 ( y 2 p)2 5 ( y 1 p)2

x2 1 y2 2 2py 1 p2 5 y2 1 2py 1 p2

x2 5 4py

Problem Solving

55. Focus is (0, 6)

p 5 6

x2 5 4py

x2 5 4(6)y 5 24y

An equation for the cross section of the trough is x2 5 24y.

Diameter 5 17 ft

Radius 5 8.5 ft

When x 5 8.5:

(8.5)2 5 24y

72.25

} 24

5 y

3.01 ø y The trough is about 3 feet deep.

56. y2 5 4px, p > 0

When x 5 3 and y 5 4:

(4)2 5 4p(3)

16 5 12p

4 }

3 5 p

The focal length is 4 }

3 inches.

57. a.

x

y

(0, 48)

146 in.

48 in.

x

y

146 in.

48 in.

(248, 0)

b. x2 5 4py y2 5 4px

x2 5 4(48)y y2 5 4(248)

x2 5 192y y2 5 2192x

c. Using x2 5 192y and x 5 146

} 2 5 73:

x2 5 192y

732 5 192y

27.8 ø y The dish is about 27.8 5 27.8 inches deep.

Using y2 5 2192 x and y 5 146

} 2 5 73:

y2 5 2192x

732 5 2192x

227.8 ø x The dish is about 227.8 5 27.8 inches deep.

No, you get the same answer using each equation.

58. Because the parabola opens up, p is positive.

Focal distance 5 p 5 p

p 5 0.36(diameter)

p 5 0.36(25) 5 9

x2 5 4py

x2 5 4(9)y 5 36y

An equation for the cross section of the dish is x2 5 36y.

When x 5 25

} 2 5 12.5:

(12.5)2 5 36y

4.3 ø y Each dish is about 4.3 meters deep.

59. a. When y 5 9.5:

x2 5 10.5y

x2 5 10.5(9.5)

x2 5 99.75

x ø 9.99

The diameter is about 2(9.99) ø 20 inches.


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b. Original refl ector: x2 5 10.5y.

For a wider refl ector, 4p > 10.5.

Sample answer:

Let p 5 12.5: x2 5 4py

x2 5 4(12.5)y

x2 5 50y

When y 5 9.5: x2 5 50(9.5)

x2 5 475

x ø 21.8

The wider refl ector’s beam is about 2(21.8) 5 43.6 inches wide.

c. For a narrower refl ector, 4 p < 10.5.

Sample answer:

Let p 5 2: x2 5 4py

x2 5 4(2)y

x2 5 8y

When y 5 9.5: x2 5 8(9.5)

x2 5 76

x ø 8.7

The narrower refl ector’s beam is about 2(8.7) 5 17.4 inches wide.

60. x2 5 4py

Find the x-values where y 5 p:

x2 5 4p2

x 5 62p

The length of the latus rectum is the distance from (22p, p) to (2p, p):

d 5 Ï}}}

(2p 2 (22p))2 1 ( p 2 p)2 5 Ï}

(4p)2 5 4p

The length of the latus rectum is 4 p.

Mixed Review

61. (x2 2 3) 2 (2x2 1 4 x 2 1) 5 x2 2 3 1 x2 2 4 x 1 1

5 2x2 2 4x 2 2

62. (2x2 2 2 x 2 3) 1 (7 2 x 2 3x3) 5 23x3 1 2x2 2 2x 2 x 2 3 1 7

5 23x3 1 2x2 2 3x 1 4

63. (x2 2 5x)(2x2 1 9x 2 3) 5 2x4 1 9x3 2 3x2 2 10x3 2 45x2 1 15x

5 2x4 2 x3 2 48x2 1 15x

64. (x 1 1)(2x 2 3)(x 2 4)

5 (2x2 2 3x 1 2x 2 3)(x 2 4)

5 (2x2 2 x 2 3)(x 2 4)

5 2x3 2 8x2 2 x2 1 4x 2 3x 1 12

5 2x3 2 9x2 1 x 1 12

65. 24x3y2

} 2y2

p x4y }

29x2y3 5 2

24x7y3

} 18 x2y5

5 2 4x5

} 3y2

66. x2 1 15x 1 36

}} 2x2 2 18

4 8x 1 96

} x 2 3 5 x2 1 15x 1 36

}} 2x2 2 18

p x 2 3 }

8x 1 96

5 (x 1 12)(x 1 3)(x 2 3)

}} 2(x 1 3)(x 2 3)8(x 1 12)

5 1 } 16

67. x }

x2 2 25 1

x 2 1 }

x2 2 3x 2 40 5

x }}

(x 1 5)(x 2 5) 1

x 2 1 }}

(x 1 5)(x 2 8)

5 x(x 2 8)

}} (x 1 5)(x 2 5)(x 2 8)

1 (x 2 1)(x 2 5)

}} (x 1 5)(x 2 5)(x 2 8)

5 x2 2 8x 1 x2 2 6x 1 5

}} (x 1 5)(x 2 5)(x 2 8)

5 2x2 2 14x 1 5

}} (x 2 5)(x 1 5)(x 2 8)

68. 6x }

x2 1 x 2 20 2

3 } x 2 4 5

6x }}

(x 2 4)(x 1 5) 2

3(x 1 5) }}

(x 2 4)(x 1 5)

5 6x 2 3x 2 15

}} (x 2 4)(x 1 5)

5 3x 2 15

}} (x 2 4)(x 1 5)

5 3(x 2 5)

}} (x 2 4)(x 1 5)

69. (2, 1), (8, 3)

d 5 Ï}}

(8 2 2)2 1 (3 2 1)2

5 Ï}

62 1 22 5 Ï}

36 1 4 5 Ï}

40 5 2 Ï}

10

70. (8, 21)(22, 4)

d 5 Ï}}}

(8 2 (22))2 1 (21 2 4)2

5 Ï}}

(10)2 1 (25)2 5 Ï}

100 1 25 5 Ï}

125 5 5 Ï}

5

71. (23, 2), (5, 1)

d 5 Ï}}

(23 2 5)2 1 (2 2 1)2

5 Ï}}

(28)2 1 (1)2 5 Ï}

64 1 1 5 Ï}

65

72. (215, 6), (21, 23)

d 5 Ï}}}

(215 2 (21))2 1 (6 2 (23))2

5 Ï}}

(214)2 1 92 5 Ï}

196 1 81 5 Ï}

277

73. (2.3, 1.1), (24.6, 21.4)

d 5 Ï}}}

(2.3 2 (24.6))2 1 (1.1 2 (21.4))2

5 Ï}

6.92 1 2.52 5 Ï}}

47.61 1 6.25 5 Ï}

53.86 ø 7.34

74. (9.4, 6.3), (8.5, 23.7)

d 5 Ï}}}

(9.4 2 8.5)2 1 (6.3 2 (23.7))2

5 Ï}

0.92 1 102 5 Ï}

0.81 1 100 5 Ï}

100.81 ø 10.04

Lesson 9.3


1. x2 1 y2 5 9 (0, 3)

(0, 23)

(3, 0)

(23, 0)

1

1 x

y

r2 5 9

r 5 3


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545Algebra 2


2. y2 5 2x2 1 49

2

x

y

22

(0, 7)

(0, 27)

(7, 0)

(27, 0)

x2 1 y2 5 49

r2 5 49

r 5 7

3. x2 2 18 5 2y2

2

2

x

y

23( ), 0

223( ), 023( )0,

223( )0,

x2 1 y2 5 18

r2 5 18

r 5 3 Ï}

2

4. The radius is the distance between the center (0, 0) and (5, 21).

r 5 Ï}}

(5 2 0)2 1 (21 2 0)2

r 5 Ï}

52 1 (21)2

r 5 Ï}

26

r2 5 26

An equaton for the circle is x2 1 y2 5 26.

5. Slope of radius to point (6, 1): m 5 1 2 0

} 6 2 0 5 1 } 6

Slope of tangent line: 2 1 } m 5 26

Equation of tangent line:

y 2 1 5 26(x 2 6)

y 2 1 5 26x 1 36

y 5 26x 1 37

6. Find the point (x , 9) where x < 0 on the circle. x2 1 y2 5 102.

x2 1 y2 5 102

x2 1 92 5 102

x2 5 19

x ø 64.4

You will be in the tower’s range from (4, 9) to (24.4, 9), a distance of 4 2 (24.4) 5 8.4 miles.


Skill Practice

1. The radius of a circle is the distance from any point on the circle to a fi xed point called the circle’s center.

2. A line tangent to a circle is perpendicular to the radius at the point of tangency, so their slopes are negative reciprocals of one another.

3. C; x2 1 y2 5 9 4. E; x2 1 y2 5 36

r2 5 9 r2 5 36

r 5 3 r 5 6

5. A; x2 1 y2 5 4 6. D; x2 1 y2 5 6

r2 5 4 r2 5 6

r 5 2 r ø 2.4

7. F; x2 1 y2 5 16 8. B; x2 1 y2 5 3

r2 5 16 r2 5 3

r 5 4 r ø 1.7

9. x2 1 y2 5 1 10. x2 1 y2 5 81

r2 5 1 r2 5 81

r 5 1 r 5 9

x

y

0.5

0.5

(1, 0)

(21, 0)

(0, 1)

(0, 21)

(0, 9)

(0, 29)

(9, 0)

(29, 0)

3

x

y

23

11. x2 1 y2 5 25 12. x2 1 y2 5 12

r2 5 25 r2 5 12

r 5 5 r 5 2 Ï}

3

(5, 0)(25, 0)

(0, 5)

(0, 25)

2

x

y

22

3, 02

3 20,

3 220,3, 022

( )

( )( )

( )1

x

y

21

13. y2 5 27 2 x2 14. x2 5 2y2 1 40

x2 1 y2 5 27 x2 1 y2 5 40

r2 5 27 r2 5 40

r 5 3 Ï}

3 r 5 2 Ï}

10

2

x

y

22

30, 3 ( )

3( )0, 233, 0( ) 23

3, 0( ) 3

2

x

y

22

10( )0, 22

10, 0( )2

10, 0( )22

10( )0, 2

15. x2 5 15 2 y2 16. y2 5 2x2 1 9

x2 1 y2 5 15 x2 1 y2 5 9

r2 5 15 r2 5 9

r 5 Ï}

15 r 5 3

1

x

y

21

15( )0,

15( )0,2 15, 0( )2

15, 0( )

1

x

y

21

(3, 0)

(23, 0)

(0, 23)

(0, 3)


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17. 15x2 1 15y2 5 60 18. 7x2 1 7y2 5 112

x2 1 y2 5 4 x2 1 y2 5 16

r2 5 4 r2 5 16

r 5 2 r 5 4

1

x

y

21

(0, 2)

(0, 22)

(2, 0)(22, 0)

(4, 0)(24, 0)

(0, 4)

(0, 24)

1

x

y

21

19. 4x2 1 4y2 5 128 20. 8x2 1 8y2 5 192

x2 1 y2 5 32 x2 1 y2 5 24

r2 5 32 r2 5 24

r 5 4 Ï}

2 r 5 2 Ï}

6

2

x

y

22

24( )0,2, 04

224( )0,

( )

2, 024( )

2

x

y

22

6, 02

6 20,

6 220,6, 022

( )

( )( )

( )

21. A; 3x2 1 3y2 5 54

x2 1 y2 5 18

r2 5 18

r 5 3 Ï}

2

22. r 5 12 23. r 5 8

x2 1 y2 5 122 x2 1 y2 5 82

x2 1 y2 5 144 x2 1 y2 5 64

24. r 5 2 25. r 5 16

x2 1 y2 5 22 x2 1 y2 5 162

x2 1 y2 5 4 x2 1 y2 5 256

26. r 5 Ï}

2 27. r 5 Ï}

15

x2 1 y2 5 ( Ï}

2 )2 x2 1 y2 5 ( Ï}

15 )2

x2 1 y2 5 2 x2 1 y2 5 15

28. r 5 5 Ï}

2 29. r 5 4 Ï}

6

x2 1 y2 5 (5 Ï}

2 )2 x2 1 y2 5 (4 Ï

}

6 )2

x2 1 y2 5 25(2) x2 1 y2 5 16(6)

x2 1 y2 5 50 x2 1 y2 5 96

30. The radius should be squared.

Center: (0, 0); Radius: 12

Equation: x2 1 y2 5 122

x2 1 y2 5 144

31. r 5 Ï}}

(26 2 0)2 1 (0 2 0)2

r 5 Ï}

36

r 5 6

x2 1 y2 5 (6)2

x2 1 y2 5 36

32. r 5 Ï}}

(0 2 0)2 1 (5 2 0)2

r 5 Ï}

25 5 5

x2 1 y2 5 52

x2 1 y2 5 25

33. r 5 Ï}}

(24 2 0)2 1 (3 2 0)2

r 5 Ï}

16 1 9 5 Ï}

25 5 5

x2 1 y2 5 52

x2 1 y2 5 25

34. r 5 Ï}}

(2 2 0)2 1 (24 2 0)2

r 5 Ï}

4 1 16 5 Ï}

20 5 2 Ï}

5

x2 1 y2 5 (2 Ï}

5 )2

x2 1 y2 5 20

35. r 5 Ï}}

(26 2 0)2 1 (8 2 0)2

r 5 Ï}

36 1 64 5 Ï}

100 5 10

x2 1 y2 5 102

x2 1 y2 5 100

36. r 5 Ï}}

(29 2 0)2 1 (2 2 0)2

r 5 Ï}

81 1 4 5 Ï}

85

x2 1 y2 5 ( Ï}

85 )2

x2 1 y2 5 85

37. r 5 Ï}}

(4 2 0)2 1 (210 2 0)2

r 5 Ï}

16 1 100 5 Ï}

116

x2 1 y2 5 ( Ï}

116 )2

x2 1 y2 5 116

38. r 5 Ï}}

(28 2 0)2 1 (25 2 0)2

r 5 Ï}

64 1 25 5 Ï}

89

x2 1 y2 5 ( Ï}

89 )2

x2 1 y2 5 89

39. r 5 Ï}}

(28 2 0)2 1 (14 – 0)2

r 5 Ï}

64 1 196 5 Ï}

260

x2 1 y2 5 ( Ï}

260 )2

x2 1 y2 5 260

40. r 5 Ï}}

(5 2 0)2 1 (212 2 0)2

r 5 Ï}

25 1 144 5 Ï}

169 5 13

x2 1 y2 5 132

x2 1 y2 5 169

41. r 5 Ï}}}

(211 2 0)2 1 (211 2 0)2

r 5 Ï}

121 1 121 5 Ï}

242

x2 1 y2 5 ( Ï}

242 )2

x2 1 y2 5 242

42. r 5 Ï}}

(9 2 0)2 1 (40 2 0)2

r 5 Ï}

81 1 1600 5 Ï}

1681 5 41

x2 1 y2 5 412

x2 1 y2 5 1681


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547Algebra 2


43. C; r 5 Ï}}

(4 2 0)2 1 (26 2 0)2

r 5 Ï}

16 1 36 5 Ï}

52

x2 1 y2 5 ( Ï}

52 )2

x2 1 y2 5 52

44. y2 1 x2 5 49 45. 4x2 1 y 5 0

r2 5 49 4x2 5 2y

r 5 7 x2 5 2 1 } 4 y

2

2 x

y(0, 7)

(0, 27)

(27, 0)(7, 0)

1

x

y

21

46. 7x2 1 7y2 5 63

1

1

x

y(0, 3)

(0, 23)

(3, 0)

(23, 0)

x2 1 y2 5 9

r2 5 9

r 5 3

47. y2 2 121 5 2x2

x

y

(0, 211)

(11, 0)

(211, 0)

(0, 11)

3

3

x2 1 y2 5 121

r2 5 121

r 5 11

48. x2 1 16y 5 0 49. 3x 5 2y2

x2 5 216y y2 5 23x

1

1

x

y

2

1 x

y

50. 12x2 1 12y2 5 192 51. 2x2 1 2y2 5 16

x2 1 y2 5 16 x2 1 y2 5 8

r2 5 16 r2 5 8

r 5 4 r 5 2 Ï}

2

(0, 4)

(0, 24)

(4, 0)(24, 0)1

1

x

y

1

1 x

y

22( ), 0

222 ), 0

22( )0,

222(( )0,

52. 6x 1 6y2 5 0

1

2

x

y

6y2 5 26x

y2 5 2x


} 1 2 0 5 4


1 } 4


y 2 4 5 2 1 } 4 (x 2 1)

y 2 4 5 2 1 } 4 x 1

1 } 4

y 5 2 1 } 4 x 1

17 }

4


} 2 2 0 5 2 3 } 2

Slope of tangent line: 2 1 } m 5

2 } 3


y 2 (23) 5 2 } 3 (x 2 2)

y 1 3 5 2 } 3 x 2

4 } 3

y 5 2 } 3 x 2

13 } 3


} 25 2 0 5 2

3 } 5


5 } 3


y 2 3 5 5 } 3 (x 2 (25))

y 2 3 5 5 } 3 x 1

25 }

3

y 5 5 } 3 x 1

34 } 3


} 26 2 0 5

1 } 3



y 2 (22) 5 23(x 2 (26))

y 1 2 5 23x 2 18

y 5 23x 2 20


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} 25 2 0 5 2

9 } 5


5 } 9


y 2 9 5 5 } 9 (x 2 (25))

y 2 9 5 5 } 9 x 1

25 } 9

y 5 5 } 9 x 1

106 } 9


} 15 2 0 5 1 } 3



y 2 5 5 23(x 2 15)

y 2 5 5 23x 1 45

y 5 23x 1 50

59. The circles must lie between the circles that pass through (23, 5) and (26, 2).

Circle that passes through (23, 5):

(23)2 1 52 5 r2

9 1 25 5 r2

Ï}

34 5 r

Circle that passes through (26, 2):

(26)2 1 22 5 r2

36 1 4 5 r2

Ï}

40 5 r

So, circles between these two circles have radii r such that 34 < r2 < 40.

Sample answers:

x2 1 y2 5 35

x2 1 y2 5 36

x2 1 y2 5 38

60. x2 1 y2 5 r2

y2 5 r2 2 x2

y 5 Ï}

r2 2 x2

Slope of segment from (2r, 0) to (x , y):

m1 5 y 2 0

} x 2 (2r)

5 y } x 1 r 5

Ï}

r2 2 x2 } x 1 r

Slope of segment from (x , y) to (r , 0):

m2 5 y 2 0

} x 2 r 5 Ï}

r2 2 x2 } x 2 r

m1m2 5 Ï}

r2 2 x2 } x 1 r p Ï

}

r2 2 x2 } x 2 r

5 r2 2 x2

}} (x 1 r)(x 2 r)

5 (r 1 x)(r 2 x)

}} (x 1 r)(x 2 r)

5 2(x 1 r)(x 2 r)

}} (x 1 r)(x 2 r)

5 21

Because m1 and m2 are negative reciprocals, the line segments meet at a right angle.

61.

x

y

(x, y)

r

Let x2 1 y2 5 r2 represent the equation of the lefthand circle. Because l bisects r, the x-coordinate of the point

(x , y) is equal to 1 }

2 r.

When x 5 1 } 2 r: When y 5

Ï}

3 } 2 r:

1 1 } 2 r 2 2 1 y2 5 r2 l 5 2y

1 }

4 r2 1 y2 5 r2 l 5 2 1 Ï

}

3 }

2 r 2

y2 5 3 } 4 r2 l 5 Ï

}

3 r

y 5 6 Ï

}

3 } 2 r

Problem Solving

62. Perimeter of tower’s range: x2 1 y2 5 152

Region covered by the tower: x2 1 y2 ≤ 152

92 1 112 ≤? 152

81 1 121 ≤? 225

202 ≤ 225 ✓

Because the point (9, 11) falls within the radius of the circle, you are in the region served by the tower.

63. Perimeter of feeding range: x2 1 y2 5 502

Feeding range: x2 1 y2 ≤ 50

(225)2 1 402 ≤? 502

625 1 1600 ≤? 2500

2225 ≤ 2500 ✓

Because the point (225, 40) is within the radius of the circle, the location is within the bats’ feeding range.


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549Algebra 2


64. D; Perimeter of delivery area: x2 1 y2 5 1002

Free delivery area: x2 1 y2 ≤ 1002

752 1 702 ≤? 1002

10,525 ÷ 10,000

Because the point (75, 70) is not within the radius of the circle, the house located at (75, 70) is located outside the free delivery area.

65. a. Points A and F :

x

y

220

20

y 5 24A B C D E F

x2 1 y2 5 225

x2 1 y2 5 100

x2 1 y2 5 25 x2 1 (24)2 5 152

x2 1 16 5 225

x2 5 209

x ø 614.5

AF 5 14.5 2 (214.5) 5 29

The plane will be in the top most layer for about 29 miles.

b. Points B and E:

x2 1 (24)2 5 102

x2 1 16 5 100

x2 5 84

x ø 69.2

BE 5 9.2 2 (29.2) 5 18.4

The plane will be above the middle layer for about 18.4 miles.

c. Points C and D:

x2 1 (24)2 5 52

x2 1 16 5 25

x2 5 9

x 5 63

CD 5 3 2 (23) 5 6

The plane will be above the lowest layer for 6 miles.

66. The tunnel’s diameter is 8 feet, so its radius is 4 feet, and an equation for its cross-section is x2 1 y2 5 42.

Half the width of the walkway is 3 feet, so let x 5 3.

32 1 y2 5 42

x

y

6 ft

4 ft

y ft(x, y)

Walkway

y2 5 16 2 9 5 7

y 5 62.65

The height of the tunnel at the center of the walkway is about 4 1 2.65 5 6.65 feet, or about 79.8 inches. Yes, a worker who is 6 feet 2 inches, or 74 inches, could walk down the center of the walkway without ducking.

67. The sprinklers should be placed so that their circular regions intersect at the edge of the flower bed, when y 5 4.5.

x2 1 (4.5)2 5 62

x

y

3

3

(x, 4.5) y � 4.5

x2 � y2 � 62

x2 5 36 2 20.25

x2 5 15.75

x ø 63.97

The distance between the sprinklers should be about 2(3.97) 5 7.9 feet.

68. a. 40

40 x

y

b. Rating of at least 6.0: x2 1 y2 5 0 Rating of at least 5.7: x2 1 y2 ≤ 225

Rating of at least 5.4: x2 1 y2 ≤ 625

Rating of at least 5.1: x2 1 y2 ≤ 1225

c. When x 5 212 and y 5 216:

(212)2 1 (216)2 5 r2

144 1 256 5 r2

400 5 r2

Because 225 < 400 < 625, the rating will be between 5.4 and 5.7.

69. 40-mile transmitter, when y 5 10: Find (x, 10) where x > 0 on the circle x2 1 102 5 402.

x2 5 1600 2 100

x2 5 1500

x ø 38.7

60-mile transmitter, when y 5 10: Find (x, 10) where x < 0 on the circle x2 1 102 5 602.

x2 5 3600 2 100

x2 5 3500

x ø 259.2

The 60-mile transmitter

x

y

20

60

(38.7, 10)

(20.8, 10)is 80 miles away from 40-mile transmitter. When you enter the 60-mile transmitter range, you will be at a point whose x-coordinate is 80 2 59.2 5 20.8.

You will be in range of both transmitters for about 38.7 2 20.8 5 17.9 miles.


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At a speed of 60 miles per hour:

d 5 rt

17.9 5 60 t

t ø 0.3 h p 60 min }

h 5 18 min

You will be in range of both transmitters for about 18 minutes.

Mixed Review

70. 3x 1 5y 5 2 3x 1 5y 5 2

2x 1 2y 5 219 3 3 23x 1 6y 5 257

11y 5 255

y 5 25

When y 5 25:

2x 1 2(25) 5 219

2x 2 10 5 219

2x 5 29

x 5 9


71. 3x 2 y 5 22

6x 1 y 5 225

9x 5 227

x 5 23

When x 5 23:

3(23) 2 y 5 22

29 2y 5 22

2y 5 7

y 5 27

The solution is: (23, 27).

72. 2x 1 5y 5 13

3x 2 4y 5 8

5x 1 y 5 21

y 5 21 2 5x

When y 5 21 2 5x:

2x 1 5(21 2 5x) 5 13

2x 1 105 2 25x 5 13

223x 5 292

x 5 4

When x 5 4:

2(4) 1 5y 5 13

5y 5 5

y 5 1


73. 2x 1 5y 2 z 5 27

x 1 3y 1 2z 5 12

2x 1 2y 5 5 → x 5 2y 2 5

When x 5 2y 2 5: 2(2y 2 5) 1 5y 2 z 5 27

4y 2 10 1 5y 2 z 5 27

9y 2 z 5 3

z 5 9y 2 3

When x 5 2y 2 5 and z 5 9y 2 3:

2y 2 5 1 3y 1 2(9y 2 3) 5 12

23y 5 23

y 5 1

When y 5 1: x 5 2(1) 2 5 5 23

When x 5 23 and y 5 1: 23 1 3(1) 1 2z 5 12

z 5 6

The solution is (23, 1, 6).

74. x 1 2y 2 z 5 10 x 1 2y 2 z 5 10

3x 2 2y 1 z 5 218 3x 2 2y 1 z 5 218

2x 1 y 1 3z 5 16 4x 5 28

x 5 22

When x 5 22: 22 1 2y 2 z 5 10

z 5 2y 2 12

When x 5 22 and z 5 2y 2 12:

2(22) 1 y 1 3(2y 2 12) 5 16

7y 2 40 5 16

y 5 8

When y 5 8: z 5 2(8) 2 12

z 5 4


75. x 1 4y 2 2z 5 2 → x 5 2 2 4y 1 2z

3x 1 y 2 z 5 23

2x 1 5y 1 z 5 27

When x 5 2 2 4y 1 2z:

3(2 2 4y 1 2z) 1 y 2 z 5 23

2(2 2 4y 1 2z) 1 5y 1 z 5 27

211y 1 5z 5 29

2(23y 1 5z) 5 2(23)

28y 5 232

y 5 4

When y 5 4: x 5 2 2 4(4) 1 2z

x 5 2z 2 14

When y 5 4 and x 5 2z 2 14:

3(2z 2 14) 1 4 2 z 5 23

5z 5 35

z 5 7

When z 5 7: x 5 2(7) 2 14 5 0



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551Algebra 2


76. F4 21

3 2GF 8 0 1

22 5 7G 5 F 32 1 2 0 2 5 4 2 7

24 2 4 0 1 10 3 1 14G 5 F 34 25 23

20 10 17G77. F2 25 0

1 8 24

2 6 1G F 24 1

9 3

7 4G

5 F 28 2 45 1 0 2 2 15 1 0

24 1 72 2 28 1 1 24 2 16

28 1 54 1 7 2 1 18 1 4G

5 F 253 213

40 9

53 24G

78. F 4 1

2 9

0 5G F 214

3

212G

Not defi ned. The number of columns in the fi rst matrix does not match the number of rows in the second matrix.

79. x2 5 60 80. x2 5 50

x 5 6 2 Ï}

15 x 5 6 5 Ï}

2

81. c2 5 152 2 122 82. c2 5 202 2 162

c2 5 225 2 144 c2 5 400 2 256

c2 5 81 c2 5 144

c 5 6 9 c 5 6 12

83. 36 5 64 2 b2 84. 100 5 324 2 b2

b2 5 64 2 36 b2 5 324 2 100

b2 5 28 b2 5 224

b 5 6 2 Ï}

7 b 5 6 4 Ï}

14

Quiz 9.1–9.3 (p. 632)

1. (4, 23), (8, 27)

d 5 Ï}}}

(4 2 8)2 1 (23 2 (27))2

5 Ï}}

(24)2 1 (4)2

5 Ï}

16 1 16 5 Ï}

32 5 4 Ï}

2

M 1 4 1 8 } 2 ,

23 1 (27) }

2 2 5 (6, 25)

2. (22, 5), (4, 9)

d 5 Ï}}

(22 2 4)2 1 (5 2 9)2

5 Ï}}

(26)2 1 (24)2

5 Ï}

36 1 16 5 Ï}

52 5 2 Ï}

13

M 1 22 1 4 } 2 ,

5 1 9 }

2 2 5 (1, 7)

3. (25, 1), (24, 8)

d 5 Ï}}}

(25 2 (24))2 1 (1 2 8)2

5 Ï}}

(21)2 1 (27)2 5 Ï}

1 1 49

5 Ï}

50 5 5 Ï}

2

M 1 25 1 (24) } 2 ,

1 1 8 }

2 2 5 1 29

} 2 , 9 }

2 2 5 (24.5, 4.5)

4. (1, 2), (7, 1)

d 5 Ï}}

(1 2 7)2 1 (2 2 1)2

5 Ï}}

(26)2 1 12 5 Ï}

36 1 1 5 Ï}

37

M 1 1 1 7 } 2 ,

2 1 1 }

2 2 5 1 4,

3 }

2 2 5 (4, 1.5)

5. (26, 25), (21, 8)

d 5 Ï}}}

(26 2 (21))2 1 (25 2 8)2

5 Ï}}

(25)2 1 (213)2 5 Ï}

25 1 169 5 Ï}

194

M 1 26 1 (21) } 2 ,

25 1 8 }

2 2 5 1 27

} 2 , 3 }

2 2 5 (23.5, 1.5)

6. (3, 22), (6, 5)

d 5 Ï}}

(3 2 6)2 1 (22 2 5)2

5 Ï}}

(23)2 1 (27)2 5 Ï}

9 1 49 5 Ï}

58

M 1 3 1 6 } 2 ,

22 1 5 }

2 2 5 1 9 } 2 ,

3 }

2 2 5 (4.5, 1.5)

7. Focus: (0, 3); p 5 3 8. Focus: (22, 0); p 5 22

x2 5 4py y2 5 4px

x2 5 4(3) y 5 12 y y2 5 4(22) x 5 28x

9. Focus: (6, 0); p 5 6 10. Focus: (0, 24); p 5 24

y2 5 4px x2 5 4py

y2 5 4(6) x 5 24x x2 5 4(24)y 5 216y

11. Focus: (0, 5); p 5 5 12. Focus: (21, 0); p5 21

x2 5 4py y2 5 4px

x2 5 4(5) y 5 20y y2 5 4(21) x 5 24x

13. x2 1 y2 5 4 14. x2 1 y2 5 64

r2 5 4 r2 5 64

r 5 2 r 5 8

1

1 x

y(0, 2)

(0, 22)

(22, 0)(2, 0)

(8, 0)(28, 0)

(0, 8)

(0, 28)

2

2

x

y


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15. x2 1 y2 5 20 16. x2 1 y2 5 75

r2 5 20 r2 5 75

r 5 2 Ï}

5 r 5 5 Ï}

3

2

2 x

y

52( ), 0

522 ), 0

52( )0,

522((

)0,

3

x

y

23

35( ), 0

325 ), 0

35( )0,

325(( )0,

17. 3x2 1 3y2 5 48 18. 6x2 1 6y2 5 108

x2 1 y2 5 16 x2 1 y2 5 18

r2 5 16 r2 5 18

r 5 4 r 5 3 Ï}

2

1

1 x

y(0, 4)

(0, 24)

(4, 0)(24, 0)

1

1

x

y

23( ), 0

223 ), 0

23( )0,

223(

(

)0,

19. Jupiter’s orbit:

r2 5 x2 1 y2

r2 5 (350)2 1 (370)2

r2 5 122,500 1 136,900

r2 5 259,400

r ø 509

Because 509 < 650, KY Cygni would contain Jupiter’s orbit.

Graphing Calculator Activity 9.3 (p. 633)

1. x2 1 y2 5 144

224 ≤ x ≤ 24, 216 ≤ y ≤ 16

Xmax 2 Xmin

}} Ymax 2 Ymin

5 48

} 32

5 3 }

2

2. x2 1 y2 5 80

215 ≤ x ≤ 15, 210 ≤ y ≤ 10

Xmax 2 Xmin

}} Ymax 2 Ymin

5 30

} 20

5 3 }

2

3. x2 1 y2 5 576

272 ≤ x ≤ 72, 248 ≤ y ≤ 48

Xmax 2 Xmin

}} Ymax 2 Ymin

5 144

} 96

5 3 }

2

4. 0.5x2 1 0.5y2 5 12 5. 7x2 1 7y2 5 105

x2 1 y2 5 24 x2 1 y2 5 15

29 ≤ x ≤ 9, 26 ≤ y ≤ 6 29 ≤ x ≤ 9, 26 ≤ y ≤ 6

Xmax 2 Xmin

}} Ymax 2 Ymin

5 18

} 12

5 3 }

2

Xmax 2 Xmin }}

Ymax 2 Ymin 5

18 }

12 5

3 }

2

6. 16x2 1 16y2 5 9

x2 1 y2 5 9 } 16

23 ≤ x ≤ 3, 22 ≤ y ≤ 2

Xmax 2 Xmin

}} Ymax 2 Ymin

5 6 }

4 5

3 }

2

Lesson 9.4


1. x2

} 16

1 y2

} 9 5 1

(24, 0)(0, 23)

1

1 x

y

(4, 0)

(0, 3)

7, 0( ) 7, 0( )2

a2 5 16 b2 5 9

Vertices: (6 Ï}

16 , 0) 5 (64, 0)

Co-vertices: (0, 6 Ï

}

9 ) 5 (0, 63)

c2 5 a2 2 b2 5 16 2 9 5 7

c 5 Ï}

7

Foci: (6 Ï}

7 , 0)

2. x2

} 36

1 y2

} 49 5 1

2

2 x

(6, 0)

(0, 7)y

(0, 27)

(26, 0)

130, ( )

130, 2 ( )

a2 5 49 b2 5 36

Vertices: (0, 67)

Co-vertices: (66, 0)

c2 5 a2 2 b2 5 49 2 36 5 13

c 5 Ï}

13

Foci: (0, 6 Ï}

13 )


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553Algebra 2


3. 25x2 1 9y2 5 225

1

1 x

(3, 0)

(0, 5)

(0, 4)

y

(0, 24)

(23, 0)

(0, 25)

25x2

} 225

1 9y2

} 225 5 1

x2

} 9 1

y2

} 25 5 1

a2 5 25 b2 5 9

Vertices: (0, 65)


c2 5 25 2 9 5 16

c 5 Ï}

16 5 4

Foci: (0, 64)

4. Vertex: (7, 0); Co-vertex: (0, 2)

a 5 7 a2 5 49

b 5 2 b2 5 4

x2

} 49

1 y2

} 4 5 1

5. Vertex: (0, 6); Co-vertex: (25, 0)

a 5 6 a2 5 36

b 5 25 b2 5 25

x2

} 25

1 y2

} 36 5 1

6. Vertex: (0, 8); Focus: (0, 23)

a 5 8 a2 5 64

c 5 23 5 3 c2 5 9

c2 5 a2 2 b2

9 5 64 2 b2

b2 5 64 2 9 5 55

x2

} 55 1 y2

} 64 5 1

7. Vertex: (25, 0); Focus: (3, 0)

a 5 25 5 5 a2 5 25

c 5 3 c2 5 9

c2 5 a2 2 b2

9 5 25 2 b2

b2 5 25 2 9 5 16

x2

} 25

1 y2

} 16 5 1

8. The major axis is vertical, with a 5 350

} 2 5 175 and

b 5 250

} 2 5 125.

x2 }

1252 1 y2

} 1752 5 1, or

x2 }

15,625 1

y2

} 30,625 5 1

The area is A 5 π(175)(125) ø 68,722 square meters.


Skill Practice

1. An ellipse is the set of all points P such that the sum of the distances between P and two fi xed points, called the foci, is a constant.

2. If the vertices of an ellipse are located at (6a, 0) or(0, 6a) and the co-vertices are located at (0, 6b) or (6b, 0), you know that c2 5 a2 2 b2, so you can determine the value of c. The foci will then be located at (6c, 0) or (0, 6c).

3. x2

} 16

1 y2

} 4 5 1

a2 5 16 a 5 4

1

1 x

y

(4, 0)(0, 2)

(24, 0)

(0, 22)

3, 02 ( )3, 022 ( )

b2 5 4 b 5 2

c2 5 16 2 4 5 12

c 5 2 Ï}

3

Vertices: (64, 0)


Foci: (62 Ï}

3 , 0)

4. x2

} 4 1

y2

} 1 5 25

(210, 0)

2

2 x

(10, 0)(0, 5)

y

(0, 25)

3, 05( )2

3, 05( )

x2

} 100

1 y2

} 25 5 1

a2 5 100 a 5 10

b2 5 25 b 5 5

c2 5 100 2 25 5 75

c 5 Ï}

75 5 5 Ï}

3

Vertices: (610, 0)


Foci: 1 65 Ï}

3 , 0 2

5. x2

} 9 1

y2

} 49 5 1

2

2 x

y

(3, 0)

(0, 7)

(23, 0)

(0, 27)

100, 2 ( )

100, 22 ( )

a2 5 49 a 5 7

b2 5 9 b 5 3

c2 5 49 2 9 5 40

c 5 2 Ï}

10

Vertices: (0, 67)


Foci: 1 0, 62 Ï}

10 2

6. x2

} 144

1 y2

} 64 5 1

x

y

3

3

(12, 0)

(212, 0)

(0, 8)

(0, 28)

(24 , 0)5

(4 , 0)5

a2 5 144 a 5 12

b2 5 64 b 5 8

c2 5 144 2 64 5 80

c 5 4 Ï}

5

Vertices: (612, 0)


Foci: 1 64 Ï}

5 , 0 2


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7. x2

} 400

1 y2

} 81 5 1

4

x

y

24

(20, 0)(0, 9)

(0, 29)

(220, 0)

2 319, 0( )

319, 0( )

a2 5 400 a 5 20

b2 5 81 b 5 9

c2 5 400 2 81 5 319

c 5 Ï}

319

Vertices: (620, 0)


Foci: (6 Ï}

319 , 0)

8. x2

} 36

1 y2

} 225 5 1

3

x

y

23

(6, 0)

(0, 15)

(26, 0)

(0, 215)

210, 3 ( )

210, 23 ( )

a2 5 225 a 5 15

b2 5 36 b 5 6

c2 5 225 2 36 5 189

c 5 3 Ï}

21

Vertices: (0, 615)


Foci: (0, 63 Ï}

21 )9. 4x2 1 y2 5 36

2

2 x

y

(3, 0)(23, 0)

(0, 6)

(0, 26)

3 30, ( )

23 30, ( )

x2

} 9 1

y2

} 36 5 1

a2 5 36 a 5 6

b2 5 9 b 5 3

c2 5 36 2 9 5 27

c 5 3 Ï}

3

Vertices: (0, 66)


Foci: (0, 63 Ï}

3 )10. 9x2 1 y2 5 9

1

x

y

22

(1, 0)

(0, 3)

(21, 0)

(0, 23)

20, 2 ( )

20, 22 ( )

x2

} 1 1

y2

} 9 5 1

a2 5 9 a 5 3

b2 5 1 b 5 1

c2 5 9 2 1 5 8

c 5 2 Ï}

2

Vertices:(0, 63)


Foci: (0, 62 Ï}

2 )11. 16x2 1 9y2 5 144

(23, 0)

1

1 x

y

(3, 0)

(0, 4)

(0, 24)

70, ( )

70, 2 ( )

x2

} 9 1

y2

} 16 5 1

a2 5 16 a 5 4

b2 5 9 b 5 3

c2 5 16 2 9 5 7

c 5 Ï}

7

Vertices: (0, 64)


Foci: (0, 6 Ï}

7 )

12. 25x2 1 49y2 5 1225

2

x22

(7, 0)

(0, 5)

(0, 25)

y

(27, 0)

2 6, 0( )

22 6, 0( )

x2

} 49

1 y2

} 25 5 1

a2 5 49 a 5 7

b2 5 25 b 5 5

c2 5 49 2 25 5 24

c 5 2 Ï}

6

Vertices: (67, 0)


Foci: (62 Ï}

6 , 0) 13. 16x2 1 25y2 5 1600

(210, 0) 2

x22

(10, 0)

(0, 8)

(6, 0)

y

(0, 28)

(26, 0)

x2

} 100

1 y2

} 64 5 1

a2 5 100 a 5 10

b2 5 64 b 5 8

c2 5 100 2 64 5 36

c 5 6

Vertices: (610, 0)


Foci: (66, 0)

14. 72x2 1 8y2 5 648

(23, 0)

(0, 29)

2

x

y

24

(3, 0)

(0, 9) 6 20, ( )

6 20, 2 ( )

x2

} 9 1

y2

} 81 5 1

a2 5 81 a 5 9

b2 5 9 b 5 3

c2 5 81 2 9 5 72

c 5 6 Ï}

2

Vertices: (0, 69)


Foci: 1 0, 66 Ï}

2 2

15. In the equation x2

} 4 1

y2

} 16 5 1, the value of a2 is 16

and corresponds to the y-axis, not the x-axis. Therefore, the vertices should be located at (0, 64) and the co-vertices should be located at (62, 0).

1

x

y

21

(0, 4)

(0, 24)

(2, 0)(22, 0)


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555Algebra 2



} 2 1

y2

} 3 5 1, a2 5 3 and a 5 Ï}

3 .

Therefore, the vertices should be located at (0, 6 Ï}

3 ) instead of (0, 63). Likewise, b2 5 2 and b 5 Ï

}

2 , so the co-vertices should be located at (6 Ï

}

2 , 0) instead of (62, 0).

30, ( )

2 30, ( )

2, 0( )

2 2, 0( )

1

1 x

y

17. Vertex: (5, 0)

Co-vertex: (0, 23)

a 5 5 a2 5 25

b 5 3 b2 5 9

x2

} 25

1 y2

} 9 5 1

18. Vertex: (0, 210)

Co-vertex: (6, 0)

a 5 10 a2 5 100

b 5 6 b2 5 36

x2

} 36

1 y2

} 100 5 1

19. Vertex: (14, 0)

Co-vertex: (0, 29)

a 5 14 a2 5 196

b 5 9 b2 5 81

x2

} 196

1 y2

} 81 5 1

20. Vertex: (0, 26)

Co-vertex: (4, 0)

a 5 6 a2 5 36

b 5 4 b2 5 16

x2

} 16

1 y2

} 36 5 1

21. Vertex: (0, 12)

Co-vertex: (11, 0)

a 5 12 a2 5 144

b 5 11 b2 5 121

x2

} 121

1 y2

} 144 5 1

22. Vertex: (20, 0)

Co-vertex: (0, 216)

a 5 20 a2 5 400

b 5 16 b2 5 256

x2 }

400 1

y2

} 256 5 1

23. Vertex: (0, 8)

Focus: (0, 6)

a 5 8 a2 5 64

c 5 6 c2 5 36

36 5 64 2 b2

b2 5 64 2 36 5 28

x2

} 28

1 y2

} 64 5 1

24. Vertex: (4, 0) 25. Vertex: (0, 9)

Focus:( Ï}

7 , 0) Focus: (0, 24 Ï}

2 ) a 5 4 a2 5 16 a 5 9 a2 5 81

c 5 Ï}

7 c2 5 7 c 5 4 Ï}

2 c2 5 32

7 5 16 2 b2 32 5 81 2 b2

b2 5 16 2 7 5 9 b2 5 81 2 32 5 49

x2

} 16

1 y2

} 9 5 1 x2

} 49

1 y2

} 81 5 1

26. Vertex: (25, 0) 27. Vertex: (0, 24)

Focus: (3, 0) Focus: (0, 22 Ï}

3 ) a 5 5 a2 5 25 a 5 4 a2 5 16

c 5 3 c2 5 9 c 5 2 Ï}

3 c2 5 12

9 5 25 2 b2 12 5 16 2 b2

b2 5 25 2 9 5 16 b2 5 16 2 12 5 4

x2

} 25

1 y2

} 16 5 1 x2

} 4 1

y2

} 16 5 1

28. Vertex: (13, 0) 29. Co-vertex: (0, Ï}

7 ) Focus: (24 Ï

}

3 , 0) Focus: (23, 0)

a 5 13 a2 5 169 b 5 Ï}

7 b2 5 7

c 5 4 Ï}

3 c2 5 48 c 5 3 c2 5 9

48 5 169 2 b2 9 5 a2 2 7

b2 5 169 2 48 5 121 16 5 a2

x2

} 169

1 y2

} 121 5 1 x2

} 16

1 y2

} 7 5 1

30. Co-vertex: (23 Ï}

5 , 0) 31. Co-vertex: (0, 25 Ï}

7 ) Focus: (0, 6) Focus: (215, 0)

b 5 3 Ï}

5 b2 5 45 b 5 5 Ï}

7 b2 5 175

c 5 6 c2 5 36 c 5 15 c2 5 225

36 5 a2 2 45 225 5 a2 2 175

81 5 a2 400 5 a2

x2

} 45

1 y2

} 81 5 1 x2

} 400

1 y2

} 175 5 1

32. Co-vertex: (0, 15) 33. Co-vertex: (2 Ï}

15 , 0) Focus: (28, 0) Focus: (0, 14)

b 5 15 b2 5 225 b 5 2 Ï}

15 b2 5 60

c 5 8 c2 5 64 c 5 14 c2 5 196

64 5 a2 2 225 196 5 a2 2 60

289 5 a2 256 5 a2

x2

} 289

1 y2

} 225 5 1 x2

} 60

1 y2

} 256 5 1


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34. Co-vertex: (232, 0) 35. B; Vertex: (0, 212)

Focus: (0, 24) Co-vertex: (28, 0)

b 5 32 b2 5 1024 a 5 12 a2 5 144

c 5 24 c2 5 576 b 5 8 b2 5 64

576 5 a2 2 1024 x2

} 64

1 y2

} 144 5 1

1600 5 a2

x2 }

1024 1

y2

} 1600 5 1

36. x2 1 y2 5 64

2

x

y

22

(0, 8)

(0, 28)

(8, 0)(28, 0)

The equation is a circle.

r2 5 64

r 5 8

37. 25x2 1 81y2 5 2025

2

x

y

22

(0, 5)

(0, 25)

(9, 0)(29, 0)

x2

} 81

1 y2

} 25 5 1

The equation is an ellipse.

38. 36y 1 x2 5 0

x

y

3

3

(0, 29)

y 5 9 x2 5 236y

The equation is a parabola.

4p 5 236

p 5 29

Focus: (0, 29)

Directrix: y 5 9

39. 65y2 5 130x

x 5 2

x

y

21

12

12

, 0( ) y2 5 2x


4p 5 2

p 5 1 } 2

Focus: 1 1 } 2 , 0 2

Directrix: x 5 2 1 } 2

40. 30x2 1 30y2 5 480

1

x

y

21

(0, 4)

(0, 24)

(4, 0)(24, 0)

x2 1 y2 5 16


r2 5 16

r 5 4

41. x2

} 75 1 4y

} 25 5 0

x

y

1

1

(0, 23)

y 5 3

x2

} 75 5 2 4 } 25 y

x2 5 212y


4p 5 212

p 5 23

Focus: (0, 23)

Directrix: y 5 3

42. 3x2

} 48

1 4y2

} 400 5 1

2

x

y

22

(0, 10)

(0, 210)

(4, 0)(24, 0)

x2

} 16

1 y2

} 100 5 1


a2 5 100 a 5 10

b2 5 16 b 5 4

43. x2

} 64

1 y2

} 64 5 4

4

x

y

24

(0, 16)

(0, 216)

(16, 0)(216, 0)

x2 1 y2 5 256


r2 5 256

r 5 16

44. 16 x2 1 10 y2 5 160

1

x

y

21

(0, 4)

(0, 24)

10, 0( )

2 10, 0( )

x2

} 10

1 y2

} 16 5 1


a2 5 16 a 5 4

b2 5 10 b 5 Ï}

10

45. x2

} 9 1

y2

} 25 5 1 x2

9y2

251 5 1

x2

9y2

91 5 1

x2

9y2

41 5 1

1

1 x

y

The conic changes from an ellipse elongated along the y-axis to a circle to an ellipse elongated along the x-axis.


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557Algebra 2


46. Sample answer: x2

} 64

1 y2

} 49 5 1

Rewrite the equation as two functions:

y2

} 49

5 1 2 x2

} 64

y2 5 49 2 49x2

} 64

y 5 6 Î} 49 2 49x2

} 64

A viewing window that does not distort the shape of the ellipse is 212 ≤ x ≤ 12 and 28 ≤ y ≤ 8. The length of

the y-axis is 2 }

3 the length of the x-axis.

47. x2

} a2

1 y2

} b2

5 1

Foci: (c, 0) and (2c, 0)

Sum of distances from (a, 0) to the two foci:

Ï}}

(a 2 c)2 1 (0 2 0)2 1 Ï}}

(a 1 c)2 1 (0 2 0)2

5 (a 2 c) 1 (a 1 c) 5 2 a

Sum of the distances from (0, b) to the two foci:

Ï}}

(c 2 0)2 1 (0 2 b)2 1 Ï}}

(c 2 0)2 1 (0 2 b)2

5 Ï}

c2 1 b2 1 Ï}

c2 1 b2 5 2 Ï}

c2 1 b2

By defi nition of an ellipse, the two sums of the distances are equal.

2a 5 2 Ï}

c2 1 b2

a 5 Ï}

c2 1 b2

a2 5 c2 1 b2

a2 2 b2 5 c2

x

y

(a, 0)(c, 0)

(2a, 0)(2c, 0)

(0, b)

(0, 2b)

Problem Solving

48. Major axis: 81 km; minor axis: 12 km

x

y

(40.5, 0)

(−40.5, 0)

(0, 6)

(0, −6)

81 km

12 km

a 5 40.5 a2 5 1640.25

b 5 6 b2 5 36

x2 }

1640.25 1

y2

} 36 5 1

A 5 π�ab 5 π�(40.5)(6) ø 763

The area of the ellipse is about 763 square kilometers.

49. Largest field:

2a 5 185; a 5 92.5

2b 5 155; b 5 77.5

x2 }

(77.5)2 1

y2

} (92.5)2

5 1

A 5 π(77.5)(92.5) ø 22,521

Smallest field:

2a 5 135; a 5 67.5

2b 5 110; b 5 55

x2 }

(55)2 1

y2

} (67.5)2

5 1

A 5 π(67.5)(55) ø 11,663

An inequality for the possible areas (in square meters) of the fields is 11,700 ≤ A ≤ 22,500.

50. 2a 5 28 2b 5 16

a 5 14 b 5 8

a2 5 196 b2 5 64

x2

} 196

1 y2

} 64 5 1

c2 5 a2 2 b2

c2 5 196 2 64 5 132

c 5 2 Ï}

33

Foci: 1 62 Ï}

33 , 0 2 Distance between the foci 5 2 1 2 Ï

}

33 2 5 4 Ï}

33

The foci are about 4 Ï}

33 centimeters apart.

51.

x

y

Sun 35.3

34.12

The sum of the two distances is 2a.

2a 5 35.3 1 0.59

a ø 17.9

The distance from the origin to a focus c is about 17.9 2 0.59 5 17.3.

c2 5 a2 2 b2

b2 5 a2 2 c2

b2 5 (17.9)2 2 (17.3)2

b2 ø 320.4 2 299.3

b2 ø 21.1

Sample answer:

An equation for the orbit is x2 }

21.1 1

y2

} 320.4 5 1.


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52.

x

y

(−225, 0) (225, 0)

A B

450 mi.

a. The sum of the distances from any point on the ellipse to each airport remains constant, in this case 600 miles.

b. The length of }

AB is 450 miles.

1 }

2 (450) 5 225 Foci: (6225, 0)

The airports’ coordinates are (6225, 0).

c. By definition of an ellipse, the plane flies 600 miles. This is equal to the length of the major axes, 2a.

2a 5 600

a 5 300

Vertices: (6300, 0)

d. c2 5 a2 2 b2

(225)2 5 (300)2 2 b2

b2 5 90,000 2 50,625

b2 5 39,375

An equation for the ellipse is x2 }

90,000 1

y2

} 39,375 5 1.

53.

x

y

(x, y)2

6

d

x

y

As the painting slides down the wall, the two triangles remain similar. So, corresponding side lengths are proportional.

x }

2 5

d } 6

2d 5 6x

d 5 3x

Using the Pythagorean Theorem:

d2 1 y2 5 62

(3x)2 1 y2 5 36

9x2 1 y2 5 36

x2

} 4 1

y2

} 36 5 1

An equation for the path of point (x, y) is x2

} 4 1

y2

} 36 5 1.

Mixed Review

54. F 3 24

22 6G 5 18 2 8 5 10

55. F 21 24

3 12 G 5 212 2 (212) 5 0

56. F 1 0 22

3 24 0

2 0 5G 5 220 1 0 1 0 2 16 2 0 2 0 5 236

57. F 26 1 8

3 0 22

21 1 5G 5 0 1 24 1 2 2 0 2 15 2 12 5 21

58. x 5 3, y 5 26 59. x 5 8, y 5 5

y 5 a } x y 5

a } x

26 5 a } 3 5 5

a } 8

218 5 a 40 5 a

y 5 2 18

} x y 5 40

} x

When x 5 24: When x 5 24:

y 5 218

} 24 5

9 } 2 y 5

40 }

24 5 210

60. x 5 215, y 5 25 61. x 5 26, y 5 3 } 4

y 5 a } x y 5

a } x

25 5 a }

215 3 }

4 5

a }

26

75 5 a 2 18

} 4 5 a

y 5 75

} x 2 9 } 2 5 a

When x 5 24: y 5 2 9 } 2x

y 5 75

} 24 5 2

75 } 4 When x 5 24:

y 5 2 9 }

2(24) 5

9 } 8

62. x 5 2 5 } 2 , y 5 27 63. x 5

2 } 9 , y 5

3 } 10

y 5 a } x y 5

a } x

27 5 a 1 2 2 } 5 2

3 }

10 5 a 1 9 }

2 2

27 1 2 5 } 2 2 5 a

6 }

90 5 a

35

} 2 5 a

1 }

15 5 a

y 5 35

} 2x y 5 1 } 15x


y 5 35 }

2(24) 5 2

35 } 8 y 5

1 }

15(24) 5 2

1 } 60


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559Algebra 2


64. y 5 3 } x

1

x

y

21

Domain: all real numbers except 0

Range: all real numbers except 0

65. y 5 2 4 } x

1x

21

y Domain: all real numbers except 0


66. y 5 5 } x 1 1

1

x22

y

x 5 21



67. y 5 22

} x 2 4

1x

21

y

x 5 4



68. y 5 6 } x 2 2

2x

22

y

y 5 22



69. y 5 2 } x 2 1 1 3

1

x21

y

y 5 3

x 5 1

Domain : all real numbers except 1


Problem Solving Workshop 9.4 (p. 640)

1. y 5 Î} 1002 2 x2

} 4

x1 5 0 y1 5 100

x2 5 20 y2 ø 99.5

x3 5 40 y3 ø 98.0

x4 5 60 y4 ø 95.4

x5 5 80 y5 ø 91.7

x6 5 100 y6 ø 86.6

x7 5 120 y7 5 80

x8 5 140 y8 ø 71.4

x9 5 160 y9 5 60

x10 5 180 y10 ø 43.6

A ø 20( y1 1 y2 1 y3 1 y4 1 y5 1 y6 1 y7 1

y8 1 y9 1 y10) ø 20(826.2) 5 16,524

An estimate for the area of the ellipse is about 4(16,524) 5 66,096 m2.

This is a better estimate; more rectangles means less area that extends beyond the ellipse.

2. Use rectangles whose widths are smaller than the previous set of rectangles to obtain a closer and closer approximation of the ellipse’s area.

3. a. 2a 5 250; a 5 125

2b 5 200; b 5 100

An equation of the ellipse is x2 }

1252 1 y2

} 1002 5 1.

b. 16x2 }

25 1

y2

} 1 5 1002

y2 5 1002 2 16x2

} 25

y 5 Î} 1002 2 16x2

} 25

Sample answer:

x1 5 0 y1 5 100

x2 5 25 y2 ø 98.0

x3 5 50 y3 ø 91.7

x4 5 75 y4 5 80

x5 5 100 y5 5 60

A ø 25 ( y1 1 y2 1 y3 1 y4 1 y5)

ø 25(429.7) ø 10,743

An estimate for the area of the ellipse is about 4(10,743) 5 42,972 square meters.


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Mixed Review of Problem Solving (p. 641)

1. a. y2 5 4px

(11.2)2 5 4p(8)

3.92 5 p

Focus: (3.92, 0)

The focus is 3.92 inches from the vertex.

b. y2 5 4(3.92)x

y2 5 15.68x

An equation for the cross section of the refl ector is y2 5 15.68x.

c.

x

y

4

20(0, 0)

(8, 11.2)

(8, 211.2)

(3.92, 0)focus

2. a. x2 1 y2 ≤ r2

x2 1 y2 ≤ 162

x2 1 y2 ≤ 256

An inequality describing the region covered by the radar is x2 1 y2 ≤ 256.

b. The coordinates of the second boat are (212, 212).

(212)2 1 (212)2 ≤? 256

288 µ 256

The boat is not in range of the radar.

c. The coordinates of the third boat are (4, 6).

x

y

(4, 6)(x, 6)

(0, 16)

(0, −16)

(16, 0)(−16, 0)

x2 1 (6)2 5 256

x2 5 220

x 5 62 Ï}

55

The boat will be in range until (22 Ï}

55 , 6).

Distance from (4, 6) to (22 Ï}

55 , 6):

Ï}}}

1 4 2 1 22 Ï}

55 2 2 2 1 (6 2 6)2 5 Ï}}

(4 1 2 Ï}

55 )2 ø 18.8

The third boat will be in radar range of the fi shing boat for about 18.8 miles.

3. a.

Sun

Mercury

4429

b. 44 1 29 5 2a

73 5 2a

36.5 5 a

c 5 a 2 29 5 36.5 2 29 5 7.5

c. c2 5 a2 2 b2

7.52 5 36.52 2 b2

b2 5 1332.25 2 56.25

b2 5 1276

b ø 35.7

An equation for Mercury’s orbit is x2 }

36.52 1 y2

} 35.72 5 1.

4.

x

y

2

2

x2 + y2 � 64

x2 + y2 � 25

x2 1 y2 5 25

r2 5 25

r 5 5

x2 1 y2 5 64

r2 5 64

r 5 8

5 < b < a < 8

Sample answer: An equation of an ellipse that lies

between x2 1 y2 5 25 and x2 1 y2 5 64 is x2

} 49

1 y2

} 36 5 1.

5. a.

x

y

1

2

B (16, 2)

O (0, 0)

A (−10, 1)

Slope of }

AO : m 5 1 2 0

} 210 2 0 5

1 }

210 5 2 1 } 10

2 1 } m 5 10

Midpoint of }

AO : 1 210 1 0 }

2 ,

1 1 0 }

2 2 5 1 25,

1 }

2 2


AO :

y 2 1 } 2 5 10(x 2 (25))

y 2 1 } 2 5 10x 1 50

y 5 10x 1 101

} 2

Slope of }

BO : m 5 2 2 0

} 16 2 0 5 1 } 8

2 1 } m 5 28


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561Algebra 2


Midpoint of }

BO : 1 16 1 0 }

2 ,

2 1 0 }

2 2 5 (8, 1)


BO :

y 2 1 5 28(x 2 8)

y 2 1 5 28x 1 64

y 5 28x 1 65


10x 1 101

} 2 5 28x 1 65

18x 5 29

} 2

x 5 29

} 36

y 5 10 1 29 }

36 2 1

101 } 2 5

527 } 9

The center of the circle is 1 29 }

36 ,

527 }

9 2 .

Distance from 1 29 }

36 ,

527 }

9 2 to (0, 0):

d 5 Ï}}

1 29 }

36 2 0 2 2 1 1 527

} 9 2 0 2 2 ø 58.6

The radius of the skid mark is about 58.6 meters.

b. v 5 Ï}

9.8vr ø Ï}}

9.8(0.7)(58.6) ø 20.05

The car was traveling about 20.05 meters per second.

c. As v increases, r increases. As v decreases,

r decreases since r 5 v2 }

9.8v .

6.

x

y

2

6O

A B(3, −2)(−2, −3)

Slope of }

AO : 5 23 2 0

} 22 2 0 5

3 } 2

Slope of tangent line at point A: 2 1

} 3 }

2 5 2

2 } 3

Slope of }

BO : 22 2 0

} 3 2 0

5 2 2 } 3

Slope of tangent line at point B: 21

} 2

2 } 3 5

3 } 2

Because the slopes of the two tangent lines are negative reciprocals, the two lines are perpendicular.

7. Parabola that passes through (22, 4):

y2 5 4px

42 5 4p(22)

22 5 p

y2 5 4(22)x

y2 5 28x

Parabola that passes through (23, 9):

y2 5 4px

x

y

8

2

(−3, 9)

(−2, 4)

(−2, y )

y2 � −8x

y2 � −27x 92 5 4p(23)

2 27

} 4 5 p

y2 5 227x

When x 5 22: y2 5 227(22)

y2 5 54

y 5 Ï}

54 ø 7.3

The parabola passes between (22, 4) and 1 22, Ï}

54 2 . Sample answer: Choose (22, 5).

y2 5 4px

52 5 4p(22)

2 25

} 8 5 p

An equation for the parabola is y2 5 4 1 2 25

} 8 2 x, or

y2 5 2 25

} 2 x.

8. Distance from (0, 0) to (16, 212):

d 5 Ï}}

(16 2 0)2 1 (212 2 0)2

5 Ï}

256 1 144

5 Ï}

400

5 20

The straight line distance is 20 miles.

9. x2 5 4py

62 5 4p(4)

36

} 16

5 p

9 }

4 5 p

Focus: 1 0, 9 }

4 2

The wire should be placed about 2.3 inches from the bottom.

Lesson 9.5

9.5 Guided Practice (p. 644)

1. x2

} 16

2 y2

} 49 5 1

4

y

22 x

2 65, 0( ) 65, 0( )

(0, 7)

(0, 27)

(4, 0)(24, 0)

a2 5 16 a 5 4

b2 5 49 b 5 7

c2 5 16 1 49 5 65

c 5 Ï}

65

The transverse axis is horizontal.

Foci: (6 Ï}

65 , 0); Vertices: (64, 0)

Asymptotes: y 5 6 7 } 4 x


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2. y2

} 36

2 x2 5 1 y

24 x

4

(1, 0)(21, 0)

(0, 6)

(0, 26)

370,( )

2 370,( )

a2 5 36 a 5 6

b2 5 1 b 5 1

c2 5 36 1 1 5 37

c 5 6 Ï}

37

The transverse axis is vertical.

Foci: (0, 6 Ï}

37 ); Vertices: (0, 66);

Asymptotes: y 5 6 6 } 1 x 5 66x

3. 4y2 2 9x2 5 36

2

y

21 x

(2, 0)(22, 0)

(0, 3)

(0, 23)

130,(

2 130,( )

) y2

} 9 2

x2

} 4 5 1

a2 5 9, a 5 3

b2 5 4, b 5 2

c2 5 9 1 4 5 13

c 5 Ï}

13

Vertices: (0, 63);

Foci: (0, 6 Ï}

13 ); Asymptotes: y 5 6 3 } 2 x

4. Foci: (63, 0) 5. Foci: (0, 610)

c 5 3, c2 5 9 c 5 10, c2 5 100

Vertices: (61, 0) Vertices: (0, 66)

a 5 1, a2 5 1 a 5 6 a2 5 36

c2 5 a2 1 b2 c2 5 a2 1 b2

9 5 1 1 b2 100 5 36 1 b2

8 5 b2 64 5 b2

x2 2 y2

} 8 5 1 y2

} 36

2 x2

} 64 5 1

6. a 5 3 cm, c 5 5 cm

b2 5 c2 2 a2

b2 5 25 2 9 5 16

b 5 4

y2

} 32 2

x2

} 42 5 1

When x 5 3: y2

} 32 2

32

} 42 5 1

y2

} 9 2

9 } 16 5 1

y2

} 9 5

25 } 16

y2 ø 14.06

y ø 23.75

The mirror has a height of about 23 2 (23.75) 5 0.75 centimeters.


Skill Practice

1. The points (22, 0) and (2, 0) in the graph at the right are the vertices of the hyperbola. The line segment joining these two points is the transverse axis.

2. All ellipse is the set of all points such that the sum of the distances between any point and the foci is a constant. A hyperbola is the set of all points such that the difference of the distances between any point and the foci is a constant.

3. x2

} 25

2 y2

} 4 5 1 4. x2 }

9 2

y2

} 36 5 1

a2 5 25 a 5 5 a2 5 9 a 5 3


b2 5 4 b 5 2 b2 5 36 b 5 6

c2 5 a2 1 b2 c2 5 9 1 36 5 45

c2 5 25 1 4 5 29 c 5 Ï}

45 5 3 Ï}

5

c 5 Ï}

29 Foci: (63 Ï}

5 , 0) Foci: (6 Ï

}

29 , 0) Asymptotes: y 5 62x


4

y

24 x

(3, 0)(23, 0)

(0, 6)

(0, 26)

5, 023( ) 5, 03( )

(5, 0)

(25, 0)

(0, 2)

(0, 22)

29, 0( )2 29, 0( ) 6

x

y

28

5. y2

} 81

2 x2

} 25 5 1 6. x2

} 144

2 y2

} 36 5 1

a2 5 81 a 5 9 a2 5 144 a 5 12


b2 5 25 b 5 5 b2 5 36 b 5 6

c2 5 81 1 25 5 106 c2 5 144 1 36 5 180

c 5 Ï}

106 c 5 6 Ï}

5

Foci: (0, 6 Ï}

106 ) Foci: (66 Ï}

5 , 0)

Asymptotes: y 5 6 9 } 5 x Asymptotes:

6

x

y

29

(5, 0)(25, 0)

(0, 9)

(0, 29)

106

106

0,( )

( )0, 2

y 5 6 6 }

12 x 5 6

1 } 2 x

(12, 0)(212, 0)

(0, 6)

(0, 26)

5, 026( ) 5, 06( )

4

y

8 x

7. y2

} 196

2 x2

} 100 5 1

(0, 14)

(0, 214)

(10, 0)(210, 0)

742

7422

( )

( )

0,

0,

8

y

24 x

a2 5 196 a 5 14

Vertices: (0, 614)

b2 5 100, b 5 10

c2 5 196 1 100 5 296

c 5 2 Ï}

74

Foci: (0, 62 Ï}

74 )

Asymptotes: y 5 6 14

} 10 x 5 6 7 } 5 x


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563Algebra 2


8. y2

} 49

2 x2

} 121

5 1

4

y

28 x

(0, 7)

(0, 27)

(11, 0)(211, 0)

1700,

1700, 2

( )

( )

a2 5 49 a 5 7

Vertices: (0, 67)

b2 5 121, b 5 11

c2 5 49 1 121 5 170

c 5 Ï}

170

Foci: (0, 6 Ï}

170 ); Asymptotes: y 5 6 7 } 11 x

9. 4x2 2 y2 5 256 10. 49x2 2 4y2 5 196

x2

} 64

2 y2

} 256 5 11 x2

} 4 2

y2

} 49 5 1

a2 5 64 a 5 8 a2 5 4 a 5 2


b2 5 256 b 5 16 b2 5 49 b 5 7

c2 5 64 1 256 5 320 c2 5 4 1 49 5 53

c 5 8 Ï}

5 c 5 Ï}

53

Foci: (68 Ï}

5 , 0) Foci: (6 Ï}

53 , 0)

Asymptotes: Asymptotes: y 5 6 7 } 2 x

y 5 6 16

} 8 x 5 62x 8

y

24 x

(2, 0)(22, 0)

(0, 7)

(0, 27)

53, 0( ) 2 53, 0( )

10

15

y

x

(0, 16)

(0, 216)

(8, 0)

(28, 0)5, 08

5, 028( )

( )

11. 9y2 2 25x2 5 225 12. 25y2 2 64x2 5 1600

y2

} 25

2 x2

} 9 5 1 y2

} 64

2 x2

} 25 5 1

a2 5 25 a 5 5 a2 5 64 a 5 8


b2 5 9 b 5 3 b2 5 25 b 5 5

c2 5 25 1 9 5 34 c2 5 64 1 25 5 89

c 5 Ï}

34 c 5 Ï}

89

Foci: (0, 6 Ï}

34 ) Foci: (0, 6 Ï}

89 )

Asymptotes: y 5 6 5 } 3 x Asymptotes: y 5 6

8 } 5 x

8

y

24 x

(0, 5)

(0, 25)

(3, 0)(23, 0)

2 340, ( )

340,( )

6

y

26 x

(0, 8)

(0, 28)

(5, 0)(25, 0)

89

2 89

( )

( )

0,

0,

13. 81x2 2 16y2 5 1296 14. 49y2 2 100x2 5 4900

x2

} 16

2 y2

} 81 5 1 y2

} 100

2 x2

} 49 5 1

a2 5 15 a 5 4 a2 5 100 a 5 10


b2 5 81 b 5 9 b2 5 49 b 5 7

c2 5 16 1 81 5 97 c2 5 100 1 49 5 149

c 5 Ï}

97 c 5 Ï}

149

Foci: (6 Ï}

97 , 0) Foci: (0, 6 Ï}

149 )

Asymptotes: y 5 6 9 } 4 x Asymptotes: y 5 6

10 } 7 x

6

y

26 x

(4, 0)(24, 0)

(0, 9)

(0, 29)

97, 0( )

2 97, 0( )

6

y

29 x

(7, 0)(27, 0)

(0, 10)

(0, 210)

149

149( )

( )

20,

0,

15. D; 45y2 2 200x2 5 1800

y2

} 40

2 x2

} 9 5 1

a2 5 40

b2 5 9

c2 5 40 1 9 5 49

c 5 7

Foci: (0, 67)

16. In the equation y2

} 36

2 x2

} 4 5 1, the hyperbola should open

up and down instead of left and right. The value of a2 is 36, so the value of a is 6 and the vertices should be located at (0, 66).

12

y

24 x


} 4 2 y2 5 4, the right side of the equation

should have been set equal to 1:

x2

} 16

2 y2

} 4 5 1

Therefore, a2 5 16 and a 5 4. The vertices are located at (64, 0). Similarly, b2 5 4 and b 5 2.

1

x

y

22


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18. Foci: (0, 64)

Vertices: (0, 62)

a 5 2 a2 5 4

c 5 4 c2 5 16

16 5 4 1 b2

12 5 b2

y2

} 4 2

x2

} 12 5 1

19. Foci: (66, 0) 20. Foci: (65, 0)


a 5 2, a2 5 4 a 5 1, a2 5 1

c 5 6, c2 5 36 c 5 5, c2 5 25

36 5 4 1 b2 25 5 1 1 b2

32 5 b2 24 5 b2

x2

} 4 2

y2

} 32 5 1 x2 2 y2

} 24 5 1

21. Foci: (0, 612) 22. Foci (610, 0)

Vertices: (0, 67) Vertices: (65 Ï}

3 , 0) a 5 7 a2 5 49 a 5 5 Ï

}

3 a2 5 75

c 5 12 c2 5 144 c 5 10 c2 5 100

144 5 49 1 b2 100 5 75 1 b2

95 5 b2 25 5 b2

y2

} 49

2 x2

} 95 5 1 x2

} 75 2 y2

} 25 5 1

23. Foci: (0, 64 Ï}

5 ) 24. Foci: (0, 63)

Vertices: (0, 64) Vertices: (0, 62 Ï}

2 ) a 5 4, a2 5 16 a 5 2 Ï

}

2 , a2 5 8

c 5 4 Ï}

5 , c2 5 80 c 5 3, c2 5 9

80 5 16 1 b2 9 5 8 1 b2

64 5 b2 1 5 b2

y2

} 16

2 x2

} 64 5 1 y2

} 8 2 x2 5 1

25. Foci: (63 Ï}

6 , 0) 26. C; Foci: (0, 66 Ï}

3 ) Verticies: (62, 0) Vertices: (0, 68)

a 5 2, a2 5 4 a 5 8, a2 5 64

c 5 3 Ï}

6 , c2 5 54 c 5 6 Ï}

3 , c2 5 108

54 5 4 1 b2 108 5 64 1 b2

50 5 b2 44 5 b2

x2

} 4 2

y2

} 50 5 1 y2

} 64

2 x2

} 44 5 1

27. x2

} 25

2 y2

} 49 5 1 9

x

y

29 (5, 0)(25, 0)

(0, 7)

(0, 27)

74, 0 74, 02( ) ( ) The equation is a hyperbola.

a2 5 25 a 5 5

b2 5 49 b 5 7

c2 5 25 1 49 5 74

Vertices: (65, 0)

Foci: 1 6 Ï}

74 , 0 2

28. y2 5 18x

x

y

2

2

x 5 292

, 092( )


4p 5 18

p 5 9 } 2

Focus: 1 9 } 2 , 0 2 ; Directrix: x 5 2

9 } 2

29. 48x2 1 12y2 5 48

1

4 x

y(0, 2)

(0, 22)

(1, 0)(21, 0)

30,( )

30, 2( )

x2

} 1 1

y2

} 4 5 1


a2 5 4 a 5 2

b2 5 1 b 5 1

c2 5 4 2 1 5 3

Foci: (0, 6 Ï}

3 ) Vertices: (0, 62); Co-vertices: (61, 0)

30. x2

} 144

1 y2

} 256 5 1

4

4 x

y

(12, 0)(212, 0)

(0, 16)

(0, 216)

0, 4 7( )

0, 24 7( )


a2 5 256 a 5 16

b2 5 144 b 5 12

c2 5 256 2 144 5 112

c 5 4 Ï}

7

Foci: (0, 64 Ï}

7 ) Vertices: (0, 616); Co-vertices: (612, 0)

31. y2

} 25

2 x2

} 121 5 1 32. 18x2 1 18y2 5 288

The equation is a hyperbola. x2 1 y2 5 16

a2 5 25 a 5 5 r2 5 16

b2 5 121 b 5 11 r 5 4

c2 5 25 1 121 5 146 The equation is a circle.

Vertices: (0, 65)

1

x

y

21

(0, 24)

(4, 0)(24, 0)

(0, 4)

Foci: 1 0, 6 Ï}

146 2


y

26 x

9

(0, 25)

(0, 5) (11, 0)

(211, 0)

33. a. x2 }

9 2

y2

} 36 5 1

a2 5 9 a 5 3

Vertices: (63, 0)

b2 5 36 b 5 6

c2 5 36 1 9 5 45

Foci: (63 Ï}

5 , 0) By changing b2 to 4, b 5 2, c2 5 13, and the foci

move to (6 Ï}

13 , 0). The vertices remain the same, but the foci change and the hyperbola becomes narrower.


n2ws-09a.indd 564 6/27/06 11:10:34 AM

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565Algebra 2


b. y2

} 16

2 x2

} 4 5 1

a2 5 16 a 5 4

Vertices: (0, 64)

b2 5 4 b 5 2

c2 5 20

Foci: 1 0, 62 Ï}

5 2 By changing b2 to 25, c2 5 41 and the foci move to

(0, 6 Ï}

41 ). The vertices are the same, but the foci change and the hyperbola becomes wider.

34. a. y2

} 15

2 x2

} 30 5 1

y2

} 15

5 1 1 x2

} 30

y2 5 15 1 x2

} 2

y 5 6 Î} 15 1 x2

} 2

b. x2

} 8.4

2 y2

} 5.5 5 1

2 y2

} 5.5 5 1 2 x2

} 8.4

y2

} 5.5 5 x2

} 8.4 2 1

y2 5 5.5x2

} 8.4 2 5.5

y 5 6 Î} 5.5x2

} 8.4

2 5.5

c. 5x2 2 7.5y2 5 12

27.5y2 5 12 2 5x2

7.5y2 5 5x2 2 12

y2 5 5x2 2 12

} 7.5

y 5 6 Î} 5x2 2 12

} 7.5

35. y 5 62x 5 6 b } a x

b }

a 5 2

b 5 2a

Sample answer:

a 5 1, a2 5 1; b 5 2, b2 5 4

x2

} 1 2

y2

} 4 5 1

a 5 2, a2 5 4; b 5 4, b2 5 16

x2

} 4 2

y2

} 16 5 1

a 5 3, a2 5 9; b 5 6, b2 5 36

x2

} 9 2

y2

} 36 5 1

All three hyperbolas approach the same asymptotes, but as a and b get larger the foci and vertices move outward.

36. Sample answer:

Choose (a, 0) for (x, y).

d1 5 Ï}}

(c 2 a)2 1 (0 2 0)2 5 c 2 a

d2 5 Ï}}

(a 2 (2c))2 1 (0 2 0)2 5 a 1 c

d2 2 d1 5 (a 1 c) 2 (c 2 a) 5 2a Because a is positive, 2a 5 2a.

37. Foci: (62, 0)

c 5 2

d2 2 d1 5 2a 5 2

a 5 1

c2 5 a2 1 b2

4 5 1 1 b2

3 5 b2

x2 2 y2

} 3 5 1

Problem Solving

38. a 5 33 c 5 56

c2 5 a2 1 b2

562 5 332 1 b2

3136 2 1089 5 b2

2047 5 b2

x2 }

1089 2

y2

} 2047 5 1

39. Vertices: 1 6 Ï

}

2 } 2 , 0 2

a2 5 2

} 4 5 1 } 2

Foci: 1 6 Ï

}

3 } 2 , 0 2

c2 5 3 } 4

c2 5 a2 1 b2

3 }

4 5

1 } 2 1 b2

3 }

4 2

2 } 4 5 b2

1 }

4 5 b2

x2

} 1 }

2 2

y2

} 1 }

4 5 1

An equation that models the hyperbola is 2x2 2 4y2 5 1.

40. a. a 5 13.3 c 5 28.2

28.22 5 13.32 1 b2

795.24 2 176.89 5 b2

618.35 5 b2

An equation of the hyperbola that models the June 21

path is x2 }

176.89 2

y2

} 618.35 5 1.


n2ws-09a.indd 565 6/27/06 11:10:45 AM

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b. a 5 4.1 c 5 25

252 5 4.12 1 b2

625 5 16.81 1 b2

608.19 5 b2

An equation of the hyperbola that models the

September 1 path is x2 }

16.81 2

y2

} 608.19 5 1.

41. a. 1 } 2 (61) 5 30.5; point A is located at (30.5, 0).

1 }

2 (170) 5 85; point B is located at (85, 240).

b. Vertices (630.5, 0)

Horizontal transverse axis

x2

} a2

2 y2

} b2 5 1

x2 }

30.52 2

y2

} b2

5 1

Using point B(85, 240):

852

} 30.52

2 (240)2

} b2

5 1

7225

} 930.25

2 1600

} b2

5 1

6294.75

} 930.25

5 1600

} b2

b2 ø 236.5

An equation that models the cross section is

x2 }

930.25 2

y2

} 236.5 5 1.

c. At the top outside edge of the roof, x 5 1 } 2 (84) 5 42:

422 }

930.25 2

y2

} 236.5 5 1

833.75

} 930.25

5 y2

} 236.5

y2 ø 212

y ø 14.6

The total height of the roof is h ø 40 1 14.6 5 54.6 feet.

42. a.

x

y

Area of outer circle

2 Area of inner circle

5 Area of walkway

πy2 2 πx2 5 600

An equation is πy2 2 πx2 5 600.

b. πy2 2 πx2 5 600

π(y2 2 x2) 5 600

y2 2 x2 5 600

} π

y2 5 x2 1 600

} π

y 5 Ï}

x2 1 600

} π

Sample answer:

x 1 3 6 9

y 5 Ï}

x2 1 600 } π

13.9 14.1 15.1 16.5

c. πy2 2 πx2 5 600

y2

} 600

} π 2

x2

} 600

} π 5 1

The equation represents a hyperbola.

a2 5 600

} π b2 5 600

} π

Vertices: 1 0, 6 Ï}

600

} π 2 Asymptotes: y 5 6

a }

b x 5 6 x

Because the radii of the walkway, x and y, must be positive, only the first quadrant of the graph represents solutions that make sense.

x

y

300

300

600p , 0( )

600p0,( )

600p0,( )

600p , 0( )2

2

d. As x and y increase, the width of the walkway, y 2 x, decreases because the area of concrete remains constant.

43. a. Foci (66, 0)

c 5 6 c2 5 36

Vertices: (64, 0)

a 5 4 a2 5 16

c2 5 a2 1 b2

36 5 16 1 b2

b2 5 36 2 16 5 20

An equation for the hyperbola is:

x2

} 16

2 y2

} 20

5 1.

b. Sample answer:

By defi nition of a hyperbola, the difference of the distances from any point to the two foci is a constant. Because the graph shown lies along the intersection points of the circular ripples, you can see that at any point, the difference of the radii always equals 8.


n2ws-09a.indd 566 6/27/06 11:10:52 AM

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567Algebra 2


44. Microphone A receives the sound 2 seconds after microphone B, so microphone A is

(2 sec) 1 1100 ft } sec 2 5 2200 feet farther away from the

elk than microphone B. The set of all points that are 2200 feet closer to B than to A is one branch of the

hyperbola x2

} a2

2 y2

} b2

5 1.

The microphones are 1 mile apart, so c 5 5280

} 2 5 2640.

d2 2 d1 5 2200

(c 1 a) 2 (c 2 a) 5 2200

2a 5 2200

a 5 1100

b2 5 c2 2 a2 5 26402 2 11002 5 5,759,600

So, the elk is located somewhere along the right branch

of the hyperbola x2 }

1,210,000 2

y2

} 5,759,600 5 1. There is not

enough information to determine its exact location.

x

y4000

4000

A B

Elk

d1d2

c 1 a c 2 a

Mixed Review

45. F 4 25

1 22G 1 F 6 12

24 24G

5 F 4 1 6 25 1 12

1 2 4 22 2 4G5 F 10 7

23 26G

46. F 6 24

21 3

28 15G 2 F 8 12

5 26

32 29G

5 F 6 2 8 24 2 12

21 2 5 3 2 (26)

28 2 32 15 2 (29)G 5 F 22 12

26 9

240 24G

47. F 24 21

0 5GF 0 21

6 5

23 2G

Not possible; the number of columns in the first matrix is not equal to the number of rows in the second matrix.

48. F 1 24

3 22GF 0 6 21

3 7 25G5 F 1(0) 1 (24)(3) 1(6) 1 (24)(7) 1(21) 1(24)(25)

3(0) 1 (22)(3) 3(6) 1 (22)(7) 3(21) 1 (22)(25)G 5 F 212 222 19

26 4 7G

49. y 5 x2 2 4x 1 7

y 5 (x2 2 4x 1 4) 1 7 2 4

y 5 (x 2 2)2 1 3

Vertex: (2, 3)

50. y 5 x2 2 2x 2 5

y 5 (x2 2 2x 1 1) 2 5 2 1

y 5 (x 2 1)2 2 6

Vertex: (1, 26)

51. y 5 x2 1 14x 1 54

y 5 (x2 1 14x 1 49) 1 54 2 49

y 5 (x 1 7)2 1 5

Vertex: (27, 5)

52. y 5 2x2 2 8x 2 20

y 5 2(x2 1 8x 1 16) 2 20 1 16

y 5 2(x 1 4)2 2 4

Vertex: (24, 24)

53. y 5 x2 1 9x 2 12

y 5 1 x2 1 9x 1 81

} 4 2 2 12 2 81

} 4

y 5 1 x 1 9 } 2 2 2 2

129 } 4

Vertex: 1 2 9 } 2 , 2

129 } 4 2

54. y 5 22x2 2 2x 2 9

y 5 22 1 x2 1 x 1 1 } 4 2 2 9 1

1 } 2

y 5 22 1 x 1 1 } 2 2 2 2

17 } 2

Vertex: 1 2 1 } 2 , 2

17 } 2 2

Quiz 9.4–9.5 (p. 648)

1. x2

} 25

1 y2

} 4 5 1

a2 5 25 a 5 5

Vertices: (65, 0) 3

x

y

21

(0, 2) (5, 0)

(0, 22)(25, 0)

21, 0( ) 2 21, 0( ) b2 5 4 b 5 2


c2 5 a2 2 b2

c2 5 25 2 4 5 21

c 5 6 Ï}

21

Foci: (6 Ï}

21 , 0)

2. x2

} 16

1 y2

} 49 5 1

2

x

y

22

(0, 27)

(24, 0)

(0, 7)

(4, 0)

(0, 2 33)

(0, 33) a2 5 49 a 5 7

Vertices: (0, 67)

b2 5 16 b 5 4


c2 5 a2 2 b2

c2 5 49 2 16 5 33

c 5 6 Ï}

33

Foci: 1 0, 6 Ï}

33 2


n2ws-09a.indd 567 6/27/06 11:11:06 AM

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3. 36x2 1 9y2 5 324

2

x

y

24

(0, 6)

(3, 0)

(0, 26)

(23, 0)

(0, 23 3)

(0, 3 3) x2

} 9 1

y2

} 36 5 1

a2 5 36 a 5 6

Vertices: (0, 66)

b2 5 9 b 5 3


c2 5 a2 2 b2

c2 5 36 2 9 5 27

c 5 63 Ï}

3

Foci: (0, 63 Ï}

3 )4. Vertex: (0, 5)

Co-vertex: (24, 0)

a 5 5 a2 5 25

b 5 4 b2 5 16

x2

} 16

1 y2

} 25 5 1

5. Vertex: (10, 0) 6. Co-vertex: (2 Ï}

15 , 0) Focus: (28, 0) Focus: (0, 25)

a 5 10 a2 5 100 b 5 Ï}

15 b2 5 15

c 5 8 c2 5 64 c 5 5 c2 5 25

c2 5 a2 2 b2 c2 5 a2 2 b2

64 5 100 2 b2 25 5 a2 2 15

b2 5 100 2 64 5 36 a2 5 25 1 15 5 40

x2

} 100

1 y2

} 36 5 1 x2

} 15

1 y2

} 40 5 1

7. y2

} 25

2 x2

} 64 5 1

x

y

3

6

(8, 0)(28, 0)

(0, 5)

(0, 25)

0, 89( )

0, 2 89( )

a2 5 25 a 5 5

Vertices: (0, 65)

b2 5 64 b 5 8

c2 5 a2 1 b2

c2 5 25 1 64 5 89

c 5 6 Ï}

89

Foci: (0, 6 Ï}

89 ) Asymptotes: y 5 6

5 } 8 x

8. 4x2 2 16y2 5 64

5, 02

5, 022

( )

( ) 6y

26 x

(4, 0)

(24, 0)

(0, 2)

(0, 22)

x2

} 16

2 y2

} 4 5 1

a2 5 16 a 5 4

Vertices: (64, 0)

b2 5 4 b 5 2

c2 5 a2 1 b2

c2 5 16 1 4 5 20

c 5 62 Ï}

5

Foci: (62 Ï}

5 , 0)

Asymptotes: y 5 6 2 } 4 x 5 6

1 } 2 x

9. 12y2 2 20x2 5 240 8

y

26(0, 22 5)

(0, 2 4 2)

(0, 2 5)

(0, 4 2)

(2 3, 0)(22 3, 0)

x

y2

} 20

2 x2

} 12 5 1

a2 5 20 a 5 2 Ï}

5

Vertices: (0, 62 Ï}

5 ) b2 5 12 b 5 2 Ï

}

3

c2 5 a2 1 b2

c2 5 20 1 12 5 32

c 5 64 Ï}

2

Foci: (0, 64 Ï}

2 )

Asymptotes: y 5 6 2 Ï

} 5 }

2 Ï}

3 x 5 6

Ï}

15 } 3 x

10. Foci: (65, 0) 11. Foci: (0, 63)


c 5 5 c2 5 25 c 5 3 c2 5 9

a 5 2 a2 5 4 a 5 1 a2 5 1

c2 5 a2 1 b2 c2 5 a2 1 b2

25 5 4 1 b2 9 5 1 1 b2

b2 5 25 2 4 5 21 b2 5 8

x2

} 4 2

y2

} 21 5 1 y2 2 x2

} 8 5 1

12. Foci: (63 Ï}

6 , 0) Vertices: (63, 0)

c 5 3 Ï}

6 c2 5 54

a 5 3 a2 5 9

c2 5 a2 1 b2

54 5 9 1 b2

b2 5 54 2 9 5 45

x2

} 9 2

y2

} 45 5 1

13. 2a 5 2.98 1 2.55

x

y

2.552.98

sun

2a 5 5.53

a 5 2.765

c 5 a 2 2.55

c 5 2.765 2 2.55

c 5 0.215

c2 5 a2 2 b2

(0.215)2 5 (2.765)2 2 b2

b2 ø 7.645 2 0.046 5 7.599

An equation for the orbit of 1 ceres is x2 }

7.645 1

y2

} 7.599 5 1.

Lesson 9.6

Investigating Algebra Activity 9.6 (p. 649)

1. No, the equations will not all be the same. The equations will vary as the distance of the fl ashlight from the wall changes.

2. By holding the fl ashlight parallel to and against the wall, the beam will form a branch of a hyperbola on the wall.


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569Algebra 2



1. (x 1 1)2 1 ( y 2 3)2 5 4

The equation represents a circle with center at (h, k) 5 (21, 3) and radius r 5 Ï

}

4 5 2.

(23, 3)

(21, 1)

(21, 5)

2

x

y

21

(21, 3) (1, 3)

2. (x 2 2)2 5 8( y 1 3)

The equation represents a parabola with vertex at (h, k) 5 (2, 23).

4p 5 8

p 5 2

The focus lies 2 units above the vertex, at (2, 21).

The directrix lies 2 units below the vertex, at y 5 25.

1x

y

21

y 5 25

(2, 21)

(2, 23)

3. (x 1 3)2 2 ( y 2 4)2

} 4 5 1

The equation represents a hyperbola with center at (h, k) 5 (23, 4).

a2 5 1 a 5 1

The vertices are 1 unit to the left and right of the centers at (24, 4) and (22, 4).

b2 5 4 b 5 2

c2 5 a2 1 b2

c2 5 1 1 4 5 5

c 5 Ï}

5 ø 2.23

The foci are about 2.23 units left and right of the center, at (25.23, 4) and (20.77, 4).

(24, 4)

(25.23, 4)

(20.77, 4)

(22, 4)

(23, 4) 1

x

y

22

4. (x 2 2)2

} 16

1 ( y 2 1)2

} 9 5 1

The equation is an ellipse with center at (h, k) 5 (2, 1).

a2 5 16 a 5 4

The vertices are 4 units left and right of the center, at (22, 1) and (6, 1).

b2 5 9 b 5 3

The co-vertices are 3 units above and below the center at (2, 4) and (2, 22).

2

x22

y

(2, 1)

(6, 1)(22, 1)(2, 4)

(2, 22)

5. Parabola with vertex (3, 21) and focus (3, 2)

x

y

2

8

(3, 2)

(3, 21)

The form of the equation is (x 2 h)2 5 4p( y 2 k) where p > 0.

The vertex is (3, 21), so h 5 3 and k 5 21.

The focus is 3 units away from the vertex, so p 5 3.

4p 5 4(3) 5 12

An equation is ( x 2 3)2 5 12( y 1 1).

6. Hyperbola with vertices at (27, 3) and (21, 3) and foci at (29, 3) and (1, 3)

x

y12

3

(29, 3)(27, 3)

(21, 3)

(24, 3)

(1, 3)

The foci lie on the transverse axis, so the transverse axis is horizontal. The form of the equation is

(x 2 h)2

} a2 2

(y 2 k)2

} b2 5 1.

The center is the midpoint of the vertices:

1 27 1 (21) }

2 ,

3 1 3 }

2 2 5 (24, 3)

h 5 24, k 5 3

The vertices are 3 units from the center, so a 5 3 and a2 5 9.

The foci are 5 units from the center, so c 5 5 and c2 5 25.

c2 5 a2 1 b2

25 5 9 1 b2

b2 5 25 2 9 5 16

An equation is (x 1 4)2

} 9 2

( y 2 3)2

} 16 5 1

7. (x 2 5)2

} 64

1 y2

} 16 5 1

h 5 5 k 5 0

The center of the ellipse is (5, 0).

Lines of symmetry: x 5 5 and y 5 0


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8. (x 1 5)2 5 8( y 2 2)

h 5 25 k 5 2

The vertex of the parabola is (25, 2).

Line of symmetry: x 5 25.

9. (x 2 1)2

} 49

2 ( y 2 2)2

} 121 5 1

h 5 1 k 5 2

The center of the hyperbola is (1, 2).


10. x2 1 y2 2 2x 1 4y 1 1 5 0

A 5 1, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(1)(1) 5 24

Because B2 2 4AC < 0, B 5 0, and A 5 C, the conic is a circle.

x2 1 y2 2 2x 1 4y 1 1 5 0

(x2 2 2x) 1 ( y2 1 4y) 5 21

(x2 2 2x 1 1) 1 ( y2 1 4y 1 4) 5 21 1 1 1 4

(x 2 1)2 1 ( y 1 2)2 5 4

h 5 1, k 5 22, r2 5 4, r 5 2

x

y1

4

(1, 22)

11. 2x2 1 y2 2 4x 2 4 5 0

A 5 2, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(2)(1) 5 28

Because AÞC and B2 2 4AC < 0, the conic is an ellipse.

2x2 1 y2 2 4x 2 4 5 0

2(x2 2 2x) 1 y2 5 4

2(x2 2 2x 1 1) 1 y2 5 4 1 2

2(x 2 1)2 1 y2 5 6

(x 2 1)2

} 3 1

y2

} 6 5 1

h 5 1, k 5 0

Center: (1, 0)

a2 5 6 a 5 Ï}

6 ø 2.45

b2 5 3 b 5 Ï}

3 ø 1.73

Vertices: (1, Ï}

6 ) and (1, 2 Ï}

6 ) Co-vertices: (1 1 Ï

}

3 , 0) and (1 2 Ï}

3 , 0)

x

y3

4

(1, )

(1, 2 )

(1 2 , 0)

(1, 0)

63

(1 1 , 0)36

12. y2 2 4y 2 2x 1 6 5 0

A 5 0, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(0)(1) 5 0

Because B2 2 4AC 5 0, the conic is a parabola.

y2 2 4y 2 2x 1 6 5 0

y2 2 4y 5 2x 2 6

y2 2 4y 1 4 5 2x 2 6 1 4

( y 2 2)2 5 2(x 2 1)

h 5 1, k 5 2

x

y

1

1

(1, 2)

( )32 , 2

4p 5 2

p 5 1 } 2

Vertex: (1, 2)

Focus: 1 3 } 2 , 2 2

13. 4x2 2 y2 2 16x 2 4y 2 4 5 0

A 5 4, B 5 0, C 5 21

B2 2 4AC 5 0 2 4(4)(21) 5 16

Because B2 2 4AC > 0, the conic is a hyperbola.

4x2 2 y2 2 16x 2 4y 2 4 5 0

4(x2 2 4x) 2 ( y2 1 4y) 5 4

4(x2 2 4x 1 4) 2 ( y2 1 4y 1 4) 5 4 1 4(4) 2 4

4(x 2 2)2 2 ( y 1 2)2 5 16

(x 2 2)2

} 4 2

( y 1 2)2

} 16 5 1

h 5 2, k 5 22

Center: (2, 22) x

y4

2

(2, 2)

(0, 22)

(2, 22)

(4, 22)

(2, 26)

a2 5 4 a 5 2

b2 5 16 b 5 4

Vertices: (0, 22) and (4, 22)

14. 4x2 2 6.25y2 2 12x 2 16 5 0

A 5 4, B 5 0, C 5 26.25

B2 2 4AC 5 0 2 4(4)(26.25) 5 100

Because B2 2 4AC > 0, the path is a hyperbola.

4x2 2 6.25y2 2 12x 2 16 5 0

4(x2 2 3x) 2 6.25y2 5 16

4 (x2 2 3x 1 1.52) 2 6.25y2 5 16 1 9

4(x 2 1.5)2 2 6.25y2 5 25

(x 2 1.5)2

} 6.25

2 y2

} 4 5 1

h 5 1.5, k 5 0

Center: (1.5, 0) 3

x

y

22

(21, 0) (4, 0)(1.5, 0)(1.5, 2)

(1.5, 22)

a2 5 6.25 a 5 2.5

b2 5 4 b 5 2

Vertices: (21, 0) and (4, 0)


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571Algebra 2



Skill Practice

1. Circles, ellipses, parabolas and hyperbolas are called conic sections because they are formed by the intersection of a plane and a double-napped cone.

2. If the discriminant of a general second-degree equation is less than 0 and B 5 0 and A 5 C, the conic is a circle. If it is less than 0 and either B Þ0 or AÞC, the conic is an ellipse. If the discriminant is equal to 0, the conic is a parabola. If it is greater than 0, the conic is a hyperbola.

3. (x 1 4)2 5 28( y 2 2)

h 5 24, k 5 2

Parabola 6

4 x

y

y 5 4

(24, 0)

(24, 2) Vertex: (24, 2)

4p 5 28

p 5 22

Focus: (24, 0)

Directrix: y 5 4

4. (x 2 2)2 1 ( y 2 7)2 5 9

2

x

y

22

(2, 7) h 5 2, k 5 7

Circle

Center: (2, 7)

Radius: Ï}

9 5 3

5. (x 2 6)2

} 25

2 ( y 1 1)2 5 1

h 5 6, k 5 21

Hyperbola

Center: (6, 21) (6, 0)

(6, 21)

(6, 22)

(1, 21)

(11, 21)21

x

y

22 a2 5 25 a 5 5

b2 5 1 b 5 1

Vertices: (1, 21) and (11, 21)

c2 5 a2 1 b2 5 25 1 1

5 26

c 5 Ï}

26 ø 5.1

Foci: (0.9, 21) and (11.1, 21)

Asymptotes: y 5 1 } 5 x 2

11 } 5 and y 5 2

1 } 5 x 1

1 } 5

6. (y 1 4)2

} 49

2 (x 1 8)2

} 9 5 1

h 5 28, k 5 24

(28, 3)

(28, 211)

(25, 24)

(211, 24)

(28, 24)

2

x

y

22 Hyperbola

Center: (28, 24)

a2 5 49 a 5 7

b2 5 9 b 5 3

Vertices: (28, 211) and (28, 3)

c2 5 a2 1 b2 5 49 1 9

5 58

c 5 Ï}

58 ø 7.6

Foci: (28, 211.6) and (28, 3.6)

Asymptotes: y 5 7 }

3 x 1

44 }

3 and y 5 2

7 } 3 x 2

68 }

3

7. (x 1 2)2

} 16

1 ( y 2 2)2

} 36 5 1

h 5 22, k 5 2

Ellipse

(2, 2)

6

4 x

y

(22, 8)

(22, 2)(26, 2)

(22, 24)

Center: (22, 2)

a2 5 36 a 5 6

b2 5 16 b 5 4

c2 5 a2 2 b2 5 36 2 16 5 20

c 5 Ï}

20 ø 4.5

Vertices: (22, 8) and (22, 24)

Co-vertices: (26, 2) and (2, 2)

Foci: (22, 6.5) and (22, 22.5)

8. (x 2 5)2 1 ( y 1 1)2 5 64

(5, 21)

2

x

y

24

h 5 5, k 5 21

Circle

Center: (5, 21)

Radius: Ï}

64 5 8

9. ( y 2 1)2 5 4(x 1 6) 8

x

y

22

x 5 27

(26, 1) (25, 1)

h 5 26, k 5 1

Parabola

Vertex: (26, 1)

4p 5 4

p 5 1

Focus: (25, 1)

Directrix: x 5 27

10. x2

} 25

1 ( y 2 2)2

} 4 5 1

(25, 2)

(0, 4)

(0, 2) (5, 2)

(0, 0)

1

x

y

21

h 5 0, k 5 2

Ellipse

Center: (0, 2)

a2 5 25, a 5 5

b2 5 4, b 5 2

c2 5 a2 2 b2 5 25 2 4 5 21

c ø 4.6

Vertices: (25, 2) and (5, 2)


Foci: (24.6, 2) and (4.6, 2)


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11. (x 1 3)2

} 9 2

(y 2 4)2

} 16 5 1

(0, 4)(26, 4)(23, 4)

(23, 8)

(23, 0)

2

4 x

y

h 5 23, k 5 4

Hyperbola

Center: (23, 4)

a2 5 9 a 5 3

b2 5 16 b 5 4

c2 5 a2 1 b2 5 9 1 16 5 25

c 5 5

Vertices: (26, 4) and (0, 4)

Foci: (28, 4) and (2, 4)

Asymptotes: y 5 4 }

3 x 1 8 and y 5 2

4 } 3 x

12. C; (x 2 4)2

} 16

1 ( y 2 1)2

} 4 5 1

h 5 4, k 5 1

Center: (4, 1)

a2 5 16 a 5 4

b2 5 4 b 5 2

The co-vertices are 2 units above and below the center, at (4, 3) and (4, 21).

13. Circle 14. Circle

Center: (25, 1) Center: (9, 21)

r 5 6, r2 5 36 r 5 2 r2 5 4

h 5 25, k 5 1 h 5 9, k 5 21

(x 1 5)2 1 ( y 2 1)2 5 36 (x 2 9)2 1 (y 1 1)2 5 4

15. Parabola 16. Parabola

Vertex: (24, 23) Vertex: (5, 3)

Focus: (1, 23) Directrix: y 5 6

h 5 24, k 5 23 h 5 5, k 5 3

The parabola opens The parabola opens down.

to the right. p 5 2(6 2 3) 5 2 3

p 5 1 2 (24) 5 5 (x 2 h)2 5 4p( y 2 k)

( y 2 k)2 5 4p(x 2 h) (x 2 5)2 5 212( y 2 3)

( y 1 3)2 5 20(x 1 4)

17. Ellipse

Vertices: (23, 4) and (5, 4)

Foci: (21, 4) and (3, 4)

Then center is the midpoint of the vertices.

1 23 1 5 }

2 ,

4 1 4 }

2 2 5 (1, 4)

h 5 1, k 5 4

The vertices are 4 units from the center, so a 5 4 and

a2 5 16.


c2 5 a2 2 b2

4 5 16 2 b2

b2 5 16 2 4 5 12

The major axis is horizontal.

(x 2 h)2

} a2 1

( y 2 k)2

} b2 5 1

(x 2 1)2

} 16

1 ( y 2 4)2

} 12 5 1

18. Ellipse

Vertices: (22, 1) and (22, 9)

Co-vertices (24, 5) and (0, 5)


1 22 1 (22) }

2 ,

1 1 9 }

2 2 5 (22, 5)

h 5 22, k 5 5


The co-vertices are 2 units from the center, so b 5 2 and b2 5 4.

The major axis is vertical.

(x 2 h)2

} b2 1

(y 2 k)2

} a2 5 1

(x 1 2)2

} 4 1

(y 2 5)2

} 16 5 1

19. Hyperbola

Vertices: (6, 23) and (6, 1)

Foci: (6, 26) and (6, 4)


1 6 1 6 }

2 ,

23 1 1 }

2 2 5 (6, 21)

h 5 6, k 5 21



c2 5 a2 1 b2

25 5 4 1 b2

b2 5 25 2 4 5 21

The transverse axis is vertical.

( y 2 k)2

} a2 2

(x 2 h)2

} b2 5 1

( y 1 1)2

} 4 2

(x 2 6)2

} 21 5 1

20. Hyperbola

Vertices: (1, 7) and (7, 7)

Foci: (21, 7) and (9, 7)


1 1 1 7 }

2 ,

7 1 7 }

2 2 5 (4, 7)

h 5 4, k 5 7



c2 5 a2 1 b2

25 5 9 1 b2

b2 5 25 2 9 5 16


(x 2 h)2

} a2 2

(y 2 k)2

} b2 5 1

(x 2 4)2

} 9 2

(y 2 7)2

} 16 5 1


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573Algebra 2


21. In writing the equation, the h and k values should be subtracted from x and y, not added. The correct

equation is (x 1 2)2

} 25

1 ( y 2 3)2

} 9 5 1.

22. (x 1 5)2

} 49

1 ( y 2 2)2

} 16 5 1

Ellipse with center (25, 2)


23. ( y 2 4)2 5 6(x 1 60)

Parabola with vertex (26, 4)

Line of symmetry: y 5 4

24. (x 2 1)2

} 36

2 ( y 2 2)

} 9 5 1

Hyperbola with center (1, 2)


25. ( y 2 5)2 2 (x 2 3)2

} 9 5 1

Hyperbola with center (3, 5)


26. (x 1 3)2 5 10( y 2 1)

Parabola with vertex (23, 1)

Line of symmetry: x 5 23

27. (x 1 2)2 1 ( y 1 1)2 5 121

Circle with center (22, 21)

Any line passing through (22, 21) is a line of symmetry.

28. 6x2 2 2y2 1 24x 1 2y 2 1 5 0

A 5 6, B 5 0, C 5 22

B2 2 4AC 5 0 2 4(6)(22) 5 48

The conic is a hyperbola, because B2 2 4AC > 0.

29. x2 1 y2 2 10x 2 6y 1 18 5 0

A 5 1, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(1)(1) 5 24

The conic is a circle because B2 2 4AC < 0, B 5 0, and A 5 C.

30. y2 210y 2 5x 1 57 5 0

A 5 0, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(0)(1) 5 0

The conic is a parabola because B2 2 4AC 5 0.

31. 4x2 1 y2 2 48x 2 14y 1 189 5 0

A 5 4, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(4)(1) 5 216

The conic is an ellipse because B2 2 4AC < 0 and A Þ C.

32. 9x2 1 4y2 1 8y 1 18x 2 41 5 0

A 5 9, B 5 0, C 5 4

B2 2 4AC 5 0 2 4(9)(4) 5 2144

The conic is an ellipse because B2 2 4AC < 0 and AÞC.

33. x2 2 18x 1 6y 1 99 5 0

A 5 1, B 5 0, C 5 0

B2 2 4AC 5 0 2 4(1)(0) 5 0


34. x2 1 y2 2 6x 1 8y 2 24 5 0

A 5 1, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(1)(1) 5 24


35. 8x2 2 9y2 2 40x 1 4y 1 145 5 0

A 5 8, B 5 0, C 5 29

B2 2 4AC 5 0 2 4(8)(29) 5 288

The conic is a hyperbola because B2 2 4AC > 0.

36. B; 4x2 1 y2 1 32x 2 10y 1 85 5 0

A 5 4, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(4)(1) 5 216


37. x2 1 y2 2 14x 1 4y 2 11 5 0

A 5 1, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(1)(1) 5 24

The conic is a circle because B2 2 4AC < 0, B 5 0 and A 5 C.

x2 1 y2 2 14x 1 4y 2 11 5 0

(x2 2 14x 1 49) 1 ( y2 1 4y 1 4) 5 11 1 49 1 4

(x 2 7)2 1 (y 1 2)2 5 64

h 5 7, k 5 22

(7, 22)

2

x

y

22

Center: (7, 22)

Radius: Ï}

64 5 8

38. x2 1 4y2 2 10x 1 16y 1 37 5 0

A 5 1, B 5 0, C 5 4

B2 2 4AC 5 0 2 4(1)(4) 5 216


x2 1 4y2 2 10x 1 16y 1 37 5 0

(x2 2 10x 1 25) 1 4( y2 1 4y 1 4) 5 237 1 25 1 16

(x 2 5)2 1 4( y 1 2)2 5 4

(x 2 5)2

} 4 1 ( y 1 2)2 5 1

h 5 5, k 5 22

(3, 22)

(5, 22) (7, 22)

1 x

y

21(5, 21)

(5, 23)

Center: (5, 22)

a2 5 4, a 5 2

b2 5 1, b 5 1

Vertices: (3, 22) and (7, 22)



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39. x2 2 16x 2 8y 1 80 5 0

A 5 1, B 5 0, C 5 0

B2 2 4AC 5 0 2 4(1)(0) 5 0

The conic is a parabola because

B2 2 4AC 5 0.

x2 2 16x 2 8y 1 80 5 0

(x2 2 16x 1 64) 5 8y 2 80 1 64

(x 2 8)2 5 8y 2 16

(x 2 8)2 5 8(y 2 2)

h 5 8, k 5 2

(8, 2)

(8, 4)

2

x

y

22 y 5 0

Vertex: (8, 2)

4p 5 8

p 5 2

Focus: (8, 4)

40. 9y2 2 x2 2 54y 1 8x 1 56 5 0

A 5 21, B 5 0, C 5 9

B2 2 4AC 5 0 2 4(21)(9) 5 36


9y2 2 x2 2 54y 1 8x 1 56 5 0

9( y2 2 6y 1 9) 2 (x2 2 8x 1 16) 5 256 1 81 2 16

9( y 2 3)2 2 (x 2 4)2 5 9

( y 2 3)2 2 (x 2 4)2

} 9 5 1

h 5 4, k 5 3

x

y

1

3

(1, 3) (4, 4)

(4, 3)(4, 2)

(7, 3)

Center: (4, 3)

a2 5 1, a 5 1

b2 5 9, b 5 3


41. 9x2 1 4y2 2 36x 2 24y 1 36 5 0

A 5 9, B 5 0, C 5 4

B2 2 4AC 5 0 2 4(9)(4) 5 2144


9x2 1 4y2 2 36x 2 24y 1 36 5 0

9(x2 2 4x 1 4) 1 4( y2 2 6y 1 9) 5 236 1 36 1 36

9(x 2 2)2 1 4(y 2 3)2 5 36

(x 2 2)2

} 4 1

( y 2 3)2

} 9 5 1

h 5 2, k 5 3

1

x

y

21

(0, 3) (4, 3)(2, 3)

(2, 6)

(2, 0)

Center: (2, 3)

a2 5 9 a 5 3

b2 5 4 b 5 2



42. y2 1 14y 1 16x 1 33 5 0

A 5 0, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(0)(1) 5 0


y2 1 14y 1 16x 1 33 5 0

( y2 1 14y 1 49) 5 216x 2 33 1 49

( y 1 7)2 5 216x 1 16

( y 1 7)2 5 216(x 2 1)

h 5 1, k 5 27 2

x22

y

x 5 5

(23, 27)

(1, 27)

Vertex: (1, 27)

4p 5 216

p 5 24

Focus: (23, 27)

43. x2 1 y2 1 16x 2 8y 1 16 5 0

A 5 1, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(1)(1) 5 24


x2 1 y2 1 16x 2 8y 1 16x 5 0

(x2 1 16x 1 64) 1 ( y2 2 8y 1 16) 5 216 1 64 1 16

(x 1 8)2 1 ( y 2 4)2 5 64

h 5 28, k 5 4

4

x

y

24

(28, 4)

Center: (28, 4)

Radius: Ï}

64 5 8

44. x2 2 4y2 1 8x 2 24y 2 24 5 0

A 5 1, B 5 0, C 5 24

B2 2 4AC 5 0 2 4(1)(24) 5 16


x2 2 4y2 1 8x 2 24y 2 24 5 0

(x2 1 8x 1 16) 2 4( y2 1 6y 1 9) 5 24 1 16 2 36

(x 1 4)2 2 4( y 1 3)2 5 4

(x 1 4)2

} 4 2 ( y 1 3)2 5 1

h 5 24, k 5 23 2

x

y

210 (24, 22)

(24, 23)(24, 24)

(26, 23)(22, 23)

Center: (24, 23)

a2 5 4, a 5 2

b2 5 1, b 5 1

Vertices: (26, 23) and (22, 23)


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575Algebra 2


45. If A 5 C, then the conic will be a circle. If AÞC but A and C have the same sign, the conic will be an ellipse. If either A 5 0 or C 5 0, the conic will be a parabola. If A and C have opposite signs, the conic will be a hyperbola.

46. y 5 a } x

xy 5 a

A 5 0, B 5 1, C 5 0

B2 2 4AC 5 1 2 4(0)(0) 5 1


47. The foci are c units above and below the center, at (h, k 1 c) and (h, k 2 c).

The asymptotes are y 5 6 a }

b x shifted horizontally h units

and vertically k units:

( y 2 k) 5 a }

b (x 2 h) ( y 2 k) 5 2

a }

b (x 2 h)

y 5 a }

b x 2

ah }

b 1 k y 5 2

a }

b x 1

ah }

b 1 k

y 5 a }

b x 1

bk 2 ah }

b y 5 2

a }

b x 1

bk 1 ah }

b

Problem Solving

48.

x

y

2

6

(0, 4)

(0, 24)

8 ft

Top circle:

Center: (0, 4)

Radius: 4

Equation: (x 2 0)2 1 ( y 2 4)2 5 42

x2 1 (y 2 4)2 5 16

Bottom circle:

Center: (0, 24)

Radius: 4

Equation: (x 2 0)2 1 (y 1 4)2 5 42

x2 1 (y 1 4) 5 16

49. x2 2 10x 1 4y 5 0

x2 2 10x 5 24y

x2 2 10x 1 25 5 24y 1 25

(x 2 5)2 5 24 1 y 2 25

} 4 2 An equation for the path of the leap is

(x 2 5)2 5 24 1 y 2 25

} 4 2 . Vertex: 1 5,

25 }

4 2

The person’s jump is 25

} 4 feet high and

2(5) 5 10 feet wide.

50. 21y2 2 210y 2 4x2 5 2441

A 5 24, B 5 0, C 5 21

B2 2 4AC 5 0 2 4(24)(21) 5 336

The shape of the path is a hyperbola because B2 2 4AC > 0.

21y2 2 210y 2 4x2 5 2441

21( y2 2 10y 1 25) 2 4x2 5 2441 1 525

21( y 2 5)2 2 4x2 5 84

( y 2 5)2

} 4 2

x2

} 21 5 1

Center: (0, 5) (0, 7)

(0, 3)

(0, 5)21, 521, 5 ( )( )2

1x

y

22

a 5 2, b 5 Ï}

21


51. a. For the hotel, h 5 100, k 5 260 and r 5 150.

You will be in range of the transmitter when (x 2 100)2 1 ( y 1 60)2 ≤ 1502.

For the café, h 5 280, k 5 270, and r 5 100. You will be in range of the transmitter when (x 1 80)2 1 ( y 1 70)2 ≤ 1002.

b. At point (0, 0):

(0 1 80)2 1 (0 1 70)2 ≤? 1002

6400 1 4900 ≤? 1002

11,300 µ 10,000

(0 2 100)2 1 (0 1 60)2 ≤? 1502

10,000 1 3600 ≤? 22,500

13,600 ≤ 22,500 ✓

(0, 0)

(100, 260)Hotel

(280, 270)Cafe

You

x

y

At the origin, you are in range of the hotel’s transmitter because its inequality is satisfi ed. You are not in range of the café’s transmitter because its inequality is not satisfi ed.

c. Because the café’s range is 100 yards and the hotel’s range is 150 yards, if the hotel’s distance from the café is less than their combined range of 250 yards, there is an overlap.

52. a. An ellipse is formed by cutting the cone-shaped tip, because the cut enters the cone diagonally and exits the other side.

b. Hyperbolas are formed by each fl at side and the cone-shaped tip, because the fl at sides are parallel to the axis of the cone.


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53. a. The intersection is not a circle when the plane crosses the point where the cones meet. It is a point.

b. The intersection is not a hyperbola when the plane crosses the point where the cones meet. It is a pair of lines.

c. The intersection is not a parabola when the plane lies along the edge of the cone. It is a line.

Mixed Review

54. 4x 1 3y 5 7 4x 1 3y 5 7

2x 1 2y 5 12 3 4 24x 1 8y 5 48

11y 5 55

y 5 5

4x 1 3(5) 5 7

4x 5 28

x 5 22


55. x 1 3y 5 22 3 (25) 25x 2 15y 5 10

5x 1 7y 5 14 5x 1 7y 5 14

28y 5 24

y 5 23

x 1 3(23) 5 22

x 5 7


56. 3x 2 2y 5 13 3 3 9x 2 6y 5 39

5x 1 3y 5 229 3 2 10x 1 6y 5 258

19x 5 219

x 5 21

3(21) 2 2y 5 13

22y 5 16

y 5 28


57. 4x 2 5y 5 214 3 2 8x 2 10y 5 228

3x 1 2y 5 24 3 5 15x 1 10y 5 120

23x 5 92

x 5 4

4(4) 2 5y 5 214

25y 5 230

y 5 6


58. 5x 1 2y 5 5 3 2 10x 1 4y 5 10

22x 1 3y 5 221 3 5 210x 1 15y 5 2105

19y 5 295

y 5 25

5x 1 2(25) 5 5

5x 5 15

x 5 3


59. 2x 1 y 5 27 3 (24) 28x 2 4y 5 28

5x 1 4y 5 21 5x 1 4y 5 21

23x 5 27

x 5 29

2(29) 1 y 5 27

y 5 11


60. ln x5y 5 ln x5 1 ln y 5 5 ln x 1 ln y

61. log3 25

} x 5 log3 25 2 log3 x

62. log7 48x3 5 log7 48 1 log7 x3 5 log7 48 1 3 log7 x

63. ln Ï

}

x }

3 5 ln Ï

}

x 2 ln 3 5 1 } 2 ln x 2 ln 3

64. 6 log5 s 2 3 log5 t 5 log5 s6 2 log5 t

3 5 log5 s6

} t3

65. log 20 1 log 125 1 2 log 2 5 log(20)(125)(22) 5 log 10,000

66. 2 ln x 2 2(ln y 1 0.25 ln y) 5 ln x2 1 ln y22 1 ln y20.5

5 ln x2 }

y2 Ï}

y

Lesson 9.7


1. x2 1 y2 5 13 and y 5 x 2 1

y2 5 13 2 x2

y 5 6 Ï}

13 2 x2

Using the calculator’s intersect feature, the solutions are (22, 23) and (3, 2).

2. x2 1 8y2 2 4 5 0 and y 5 2x 1 7

8y2 5 4 2 x2

y2 5 1 } 2 2

x2

} 8

y 5 6 Î} 1 }

2 2

x2

} 8

The graphs do not intersect. There is no solution.

3. y2 1 6x 2 1 5 0 and y 5 20.4x 1 2.6

y2 5 26x 1 1

y 5 6 Ï}

26x 1 1

Using the calculator’s intersect feature, the solutions are approximately (21.57, 3.23) and (222.9, 11.8).

4. y 5 0.5x 2 3

x2 1 4y2 2 4 5 0

x2 1 4(0.5x 2 3)2 2 4 5 0

x2 1 4(0.25x2 2 3x 1 9) 2 4 5 0

x2 1 x2 2 12x 1 36 2 4 5 0

2x2 2 12x 1 32 5 0

2(x2 2 6x 1 16) 5 0

There is no solution.


n2ws-09b.indd 576 6/28/06 1:41:02 PM

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577Algebra 2


5. y2 2 2x 2 10 5 0

y 5 2x 2 1

(2x 2 1)2 2 2x 2 10 5 0

x2 1 2x 1 1 2 2x 2 10 5 0

x2 2 9 5 0

(x 1 3)(x 2 3) 5 0

x 5 63


y 5 2(3) 2 1 y 5 2(23) 2 1

y 5 24 y 5 2

The solutions are (3, 24) and (23, 2).

6. y 5 4x 2 8

9x2 2 y2 2 36 5 0

9x2 2 (4x 2 8)2 2 36 5 0

9x2 2 (16x2 2 64x 1 64) 2 36 5 0

9x2 2 16x2 1 64x 2 64 2 36 5 0

27x2 1 64x 2 100 5 0

2(7x2 2 64x 1 100) 5 0

2(7x 2 50)(x 2 2) 5 0

7x 2 50 5 0 or x 2 2 5 0

x 5 50

} 7 or x 5 2

When x 5 50

} 7 : When x 5 2:

y 5 4 1 50 } 7 2 2 8 y 5 4(2) 2 8

y 5 144

} 7 y 5 0

The solutions are (2, 0) and 1 50 } 7 ,

144 } 7 2 .

7. 22y2 1 x 1 2 5 0

x2 1 y2 2 1 5 0 3 2

22y2 1 x 1 2 5 0

2x2 1 2y2 2 2 5 0

2x2 1 x 5 0

x(2x 1 1) 5 0

x 5 0 or 2x 1 1 5 0

x 5 2 1 } 2

When x 5 0: When x 5 2 1 } 2 :

02 1 y2 2 1 5 0 1 1 } 2 2 2 1 y2 2 1 5 0

y2 5 1 y2 5 3 } 4

y 61 y 5 6 Î}

3 }

2

The solutions are (0, 61) and 1 2 1 } 2 , 6

Ï}

3 } 2 2 .

8. x2 1 y2 2 16x 1 39 5 0

x2 2 y2 2 9 5 0

2x2 2 16x 1 30 5 0

(2x 2 6)(x 2 5) 5 0

x 5 3 or x 5 5


32 2 y2 2 9 5 0 52 2 y2 2 9 5 0

y2 5 0 y2 5 16

y 5 0 y 5 64


9. x2 1 4y2 1 4x 1 8y 5 88

y2 2 x 1 2y 5 5 3 (24)

x2 1 4y2 1 4x 1 8y 5 8 24y2 1 4x 2 8y 5 220

x2 1 8x 5 212

x2 1 8x 1 12 5 0

(x 1 6)(x 1 2) 5 0

x 1 6 5 0 or x 1 2 5 0

x 5 26 or x 5 22


y2 2 (26) 1 2y 5 5 y2 2 (22) 1 2y 5 5

y2 1 2y 1 1 5 0 y2 1 2y 2 3 5 0

( y 1 1) 5 0 ( y 1 3)( y 2 1) 5 0

y 5 21 y 1 3 5 0 or y 2 1 5 0

y 5 23 or y 5 1

The solutions are (26, 21), (22, 23), and (22, 1).

10. x2 2 y2 2 12x 1 18 5 0

2x2 1 y2 2 4y 1 2 5 0

212x 2 4y 1 20 5 0

24y 5 12x 2 20

y 5 23x 1 5

x2 2 (23x 1 5)2 2 12x 1 18 5 0

x2 2 (9x2 2 30x 1 25) 2 12x 1 18 5 0

28x2 1 18x 2 7 5 0

2(8x2 2 18x 1 7) 5 0

2(4x 2 7)(2x 2 1) 5 0

4x 2 7 5 0 or 2x 2 1 5 0

x 5 7 } 4 or x 5

1 } 2

When x 5 7 } 4 : When x 5

1 } 2 :

y 5 23 1 7 } 4 2 1 5 y 5 23 1 1 }

2 2 1 5

y 5 2 1 } 4 y 5

7 } 2

The solutions are 1 7 } 4 , 2

1 } 4 2 and 1 1 }

2 ,

7 }

2 2 . Because the

ship is south of the x-axis, it is at 1 7 } 4 , 2

1 } 4 2 .


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Skill Practice

1. The equations x2 1 5x 1 3y2 5 9 and 4x2 2 12y 1 16 5 0 form a quadratic system of equations.

2. 3x2 1 y2 2 5x 5 0

2x2 1 y2 2 15 5 0

Elimination could be used to solve the system because subtracting the two equations would eliminate the y2-term.

3. x2 1 y2 2 32 5 0 and y 2 x 5 0

y2 5 32 2 x2 y 5 x

y 5 6 Ï}

32 2 x2

Using the calculator’s intersect feature, the solutions are (24, 24) and (4, 4).

4. y 1 2x2 2 9 5 0 and y 1 4x 1 1 5 0

y 5 9 2 2x2 and y 5 24x 2 1


5. y 2 3x 1 4 5 0 and 23x2 1 y2 2 6 5 0

y 5 3x 2 4 and y2 5 3x2 1 6

y 5 6 Ï}

3x2 1 6


6. y 1 2x 5 6 and 3x2 1 y2 5 12

y 5 6 2 2x and y2 5 12 2 3x2

y 5 6 Ï}

12 2 3x2

The graphs do not intersect. There is no solution.

7. x2 1 y2 5 16 and y 2 2x 5 1

y2 5 16 2 x2 and y 5 2x 1 1

y 5 6 Ï}

16 2 x2


8. 3( y 1 3)2 1 4x 5 0 and y 2 2x 5 11

( y 1 3)2 5 2 4x

} 3 and y 5 2x 1 11

y 1 3 5 6 Ï}

2 4x

} 3

y 5 236 Î} 2 4x

} 3


9. y2 2 x 2 6 5 0

y 1 x 5 0 → y 5 2x

(2x)2 2 x 2 6 5 0

x2 2 x 2 6 5 0

(x 1 2)(x 2 3) 5 0

x 5 22 or x 5 3


y 5 2(22) 5 2 y 5 2(3) 5 23


10. x2 1 y2 2 25 5 0

y 5 2x 2 10

x2 1 (2x 2 10)2 2 25 5 0

x2 1 4x2 2 40x 1 100 2 25 5 0

5x2 2 40x 1 75 5 0

5(x2 2 8x 1 15) 5 0

5(x 2 3)(x 2 5) 5 0

x 5 3 or x 5 5


y 5 2(3) 2 10 5 24 y 5 2(5) 2 10 5 0


11. 22x 1 y 2 8 5 0 → y 5 2x 1 8

x2 1 4y2 2 40 5 0

x2 1 4(2x 1 8)2 2 40 5 0

x2 1 4 1 4x2 1 32x 1 64 2 2 40 5 0

x2 1 16x2 1 128x 1 256 2 40 5 0

17x2 1 128x 1 216 5 0

x 5 2128 6 Ï

}}

1282 2 4(17)(216) }}}

2(17)

x ø 22.55 or x ø 24.98

When x 5 22.55: When x 5 24.98:

y 5 2(22.55) 1 8 y 5 2(24.98) 1 8

y 5 2.9 y 5 21.96

The solutions are approximately (22.55, 2.9) and (24.98, 21.96).

12. 2x2 1 2y2 5 8

2x 1 y 5 22 → y 5 x 2 2

2x 1 2(x 2 2)2 5 8

2x2 1 2 1 x2 2 4x 1 4 2 2 8 5 0

2x2 1 2x2 2 8x 1 8 2 8 5 0

x2 2 8x 5 0

x(x 2 8) 5 0

x 5 0 or x 5 8


y 5 0 2 2 5 22 y 5 8 2 2 5 6


13. 6x2 1 3y2 5 12

y 5 2x 1 2

6x2 1 3(2x 1 2)2 5 12

6x2 1 3(x2 2 4x 1 4) 2 12 5 0

6x2 1 3x2 2 12x 1 12 2 12 5 0

9x2 2 12x 5 0

3x(3x 2 4) 5 0

x 5 0 or 3x 5 4

x 5 4 } 3

When x 5 0: When x 5 4 } 3 :

y 5 2(0) 1 2 5 2 y 5 2 1 4 } 3 2 1 2 5 2 } 3

The solutions are (0, 2) and 1 4 } 3 ,

2 }

3 2 .


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579Algebra 2


14. 23x 1 y 5 6 → y 5 3x 1 6

8x 1 y2 1 24 5 0

8x 1 (3x 1 6)2 1 24 5 0

8x 1 (9x2 1 36x 1 36) 1 24 5 0

9x2 1 44x 1 60 5 0

x 5 244 6 Ï

}}

442 2 4(9)(60) }}

2(9)

x 5 244 6 Ï

}

2224 }} 18

There is no real solution.

15. 4x2 2 5y2 5 276

2x 1 y 5 26 → y 5 22x 2 6

4x2 2 5(22x 2 6)2 5 276

4x2 2 5(4x2 1 24x 1 36) 1 76 5 0

4x2 2 20x2 2 120x 2 104 5 0

28(2x2 1 15x 1 13) 5 0

28(2x 1 13)(x 1 1) 5 0

2x 5 213 or x 5 21

x 5 2 13

} 2

When x 5 2 13

} 2 : When x 5 21:

y 5 22 1 2 13

} 2 2 2 6 y 5 22(21) 2 6

y 5 7 y 5 24

The solutions are 1 2 13

} 2 , 7 2 and (21, 24).

16. x2 1 y2 5 20

y 5 x 2 4

x2 1 (x 2 4)2 5 20

x2 1 x2 2 8x 1 16 2 20 5 0

2x2 2 8x 2 4 5 0

x 5 2(28) 6 Ï

}}

(28)2 2 4(2)(24) }}}

2(2)

x 5 8 6 Ï

}

64 1 32 }} 4 5 2 6 Ï

}

6

When x 5 2 1 Ï}

6 : When x 5 2 2 Ï}

6 :

y 5 2 1 Ï}

6 2 4 y 5 2 2 Ï}

6 2 4

y 5 22 1 Ï}

6 y 5 22 2 Ï}

6

The solutions are (2 1 Ï}

6 , 22 1 Ï}

6 ) and (2 2 Ï

}

6 , 22 2 Ï}

6 ). 17. 9x2 1 4y2 5 36

2x 1 y 5 24 → y 5 x 2 4

9x2 1 4(x 2 4)2 5 36

9x2 1 4(x2 2 8x 1 16) 2 36 5 0

9x2 1 4x2 2 32x 1 64 2 36 5 0

13x2 2 32x 1 28 5 0

x 5 2(232) 6 Ï

}}

(232)2 2 4(13)(28) }}}

2(13)

x 5 32 6 Ï

}

2432 } 26


18. x2 1 6x 1 4y 2 3 5 0

y 1 3x 1 1 5 0 → y 5 23x 2 1

x2 1 6x 1 4(23x 2 1) 2 3 5 0

x2 1 6x 2 12x 2 4 2 3 5 0

x2 2 6x 2 7 5 0

(x 1 1)(x 2 7) 5 0

x 5 21 or x 5 7


y 5 23(21) 2 1 5 2 y 5 23(7) 2 1 5 222


19. 4x2 1 2y2 2 x 2 y 5 6

3x 2 y 5 2 → y 5 3x 2 2

4x2 1 2(3x 2 2)2 2 x 2 (3x 2 2) 5 6

4x2 1 2(9x2 2 12x 1 4) 2 x 2 3x 1 2 2 6 5 0

4x2 1 18x2 2 24x 1 8 2 x 2 3x 2 4 5 0

22x2 2 28x 1 4 5 0

x 5 2(228) 6 Ï

}}

(228)2 2 4(22)(4) }}}

2(22)

x 5 7 6 3 Ï

}

3 } 11

x ø 1.11 or x ø 0.164

When x 5 1.11: When x 5 0.164:

y 5 3(1.11) 2 2 y 5 3(0.164) 2 2

y 5 1.33 y 5 21.508

The solutions are approximately (1.11, 1.33) and (0.164, 21.508).

20. 4x2 2 y2 2 32x 2 2y 5 259

2x 1 y 2 7 5 0 → y 5 22x 1 7

4x2 2 (22x 1 7)2 2 32x 2 2(22x 1 7) 5 259

4x2 2 (4x2 2 28x 1 49) 2 32x 1 4x 2 14 1 59 5 0

4x2 2 4x2 1 28x 2 49 2 32x 1 4x 1 45 5 0

24 5 0

There is no solution.

21. B; 6x2 2 5x 1 8y2 1 y 5 23

2x 1 y 5 21 → y 5 x 2 1

6x2 2 5x 1 8(x 2 1)2 1 (x 2 1) 2 23 5 0

6x2 2 5x 1 8(x2 2 2x 1 1) 1 x 2 1 2 23 5 0

6x2 2 5x 1 8x2 2 16x 1 8 1 x 2 24 5 0

14x2 2 20x 2 16 5 0

(7x 1 4)(2x 2 4) 5 0

7x 1 4 5 0 or 2x 2 4 5 0

x 5 2 4 } 7 or x 5 2

When x 5 2 4 } 7 : When x 5 2:

y 5 2 4 } 7 2 1 5 2

11 } 7 y 5 2 2 1 5 1

The solutions are 1 2 4 } 7 , 2

11 } 7 2 and (2, 1).


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22. 6x2 2 y2 2 15 5 0

x2 1 y2 2 13 5 0

7x2 2 28 5 0

7x2 5 28

x2 5 4

x 5 62


22 1 y2 2 13 5 0 (22)2 1 y2 2 13 5 0

y2 5 9 y2 5 9

y 5 63 y 5 63


23. 5x2 1 25y2 2 125 5 0

2x 1 y2 2 5 5 0 3 (225)

5x2 1 25y2 2 125 5 0

25x 2 25y2 1 125 5 0

5x2 1 25x 5 0

5x(x 1 5) 5 0

x 5 0 or x 5 25


2(0) 1 y2 2 5 5 0 2(25) 1 y2 2 5 5 0

y2 5 5 y2 5 0

y 5 6 Ï}

5 y 5 0

The solutions are 1 0, 6 Ï}

5 2 and (25, 0).

24. 10y 5 x2 2x2 1 10y 5 0

x2 2 6 5 22 x2 5 4

10y 5 4

y 5 2 } 5

When y 5 2 } 5 :

10 1 2 } 5 2 5 x2

x2 5 4

x 5 62

The solutions are 1 62, 2 } 5 2 .

25. x2 2 y2 2 4x 1 2 5 0

2x2 1 y2 2 4y 1 2 5 0

24x 2 4y 1 4 5 0

24x 5 4y 2 4

x 5 2y 1 1

When x 5 2y 1 1:

2(2y 1 1)2 1 y2 2 4y 1 2 5 0

2(y2 2 2y 1 1) 1 y2 2 4y 1 2 5 0

22y 1 1 5 0

22y 5 21

y 5 1 } 2

When y 5 1 } 2 :

x 5 2 1 1 } 2 2 1 1 5 1 } 2

The solution is 1 1 } 2 ,

1 }

2 2 .

26. x2 2 2y 5 6 x2 2 2y 5 6

x2 2 y2 5 227 3 (21) 2x2 1 y2 5 27

y2 2 2y 5 33

y2 2 2y 2 33 5 0

y 5 2(22) 6 Ï

}}

(22)2 2 4(1)(233) }}}

2(1)

y 5 1 6 Ï}

34

y ø 6.83, 24.83

When y 5 6.83: When y 5 24.83:

x2 2 2(6.83) 5 6 x2 2 2(24.83) 5 6

x2 5 19.66 x2 5 23.66

x ø 64.43 No real solution

The solutions are approximately (64.43, 6.83).

27. x2 1 2y2 2 10 5 0 3 (22) 22x2 2 4y2 1 20 5 0

4y2 1 x 1 4 5 0 4y2 1 x 1 4 5 0

22x2 1 x 1 24 5 0

x 5 2(1) 6 Ï

}}

(1)2 2 4(22)(24) }}}

2(22)

x 5 21 6 Ï

}

2193 }}

24

4y2 1 x 1 4 5 0 → y2 5 2x 2 4

} 24

y2 5 1 21 6 Ï

}

193 }

4 2 4 2

}} 4

y2 5 217 6 Ï

}

193 } 16 < 0


28. x2 1 y2 2 16x 1 39 5 0

x22 y2 2 9 5 0

2x2 2 16x 1 30 5 0

2(x2 2 8x 1 15) 5 0

2(x 2 3)(x 2 5) 5 0

x 5 3 or x 5 5


32 2 y2 2 9 5 0 52 2 y2 2 9 5 0

y2 5 0 y2 5 16

y 5 0 y 5 64



n2ws-09b.indd 580 6/27/06 11:15:21 AM

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581Algebra 2


29. x2 2 y2 2 8x 1 8y 5 24

x2 1 y2 2 8x 2 8y 5 224

2x2 2 16x 5 0

2x(x 2 8) 5 0

x 5 0 or x 5 8

When x 5 0:

(0)2 2 y2 2 8(0) 1 8y 5 24

2y2 1 8y 2 24 5 0

y 5 28 6 Ï

}}

82 2 4(21)(224) }}

2(21)

y 5 28 6 Ï

}

232 }

(22)

No real solution

When x 5 8:

(8)2 2 y2 2 8(8) 1 8y 5 24

2y2 1 8y 2 24 5 0

y 5 28 6 Ï

}}

82 24(21)(224) }}

2( 21)

y 5 28 6 Ï

}

232 }

22

No real solution

30. 16x2 2 y2 1 16y 2 128 5 0

y2 2 48x 2 16y 2 32 5 0

16x2 2 48x 2 160 5 0

16(x2 2 3x 2 10) 5 0

16(x 1 2)(x 2 5) 5 0

x 5 22 or x 5 5

When x 5 22:

16(22)2 2 y2 1 16y 2 128 5 0

2y2 1 16y 2 64 5 0

2( y2 2 16y 1 64) 5 0

2( y 2 8)( y 2 8) 5 0

y 5 8

When x 5 5

16(5)2 2 y2 1 16y 2 128 5 0

2y2 1 16y 1 272 5 0

y 5 216 6 Ï

}}

162 2 4(21)(272) }}}

2(21)

y 5 8 6 4 Ï}

21

The solutions are (22, 8) and (5, 8 6 4 Ï}

21 ).

31. 4x2 2 56x 1 9y2 5 2160

4x2 1 y2 2 64 5 0 3 (29)

4x2 2 56x 1 9y2 1 160 5 0

236x2 2 9y2 1 576 5 0

232x2 2 56x 1 736 5 0

28(4x2 1 7x 2 92) 5 0

28(4x 1 23)(x 2 4) 5 0

4x 1 23 5 0 or x 2 4 5 0

x 5 2 23

} 4 or x 5 4

When x 5 2 23

} 4 : When x 5 4:

4 1 2 23

} 4 2 2 1 y2 2 64 5 0 4(4)2 1 y2 2 64 5 0

529

} 4 1 y2 2 64 5 0 64 1 y2 2 64 5 0

y2 5 2 273

} 4 y2 5 0

No real solution y 5 0


32. x2 2 y2 2 32x 1 128 5 0

2x2 1 y2 2 8y 1 8 5 0

232x 2 8y 1 136 5 0

28y 5 32x 2 136

y 5 24x 1 17

When y 5 24x 1 17:

x2 2 (24x 1 17)2 2 32x 1 128 5 0

x2 2 (16x2 2 136x 1 289) 2 32x 1 128 5 0

215x2 1 104x 2 161 5 0

2(15x2 2 104x 1 161) 5 0

2(5x 2 23)(3x 2 7) 5 0

5x 5 23 or 3x 5 7

x 5 23

} 5 or x 5 7 } 3

When x 5 23

} 5 : When x 5 7 } 3 :

y 5 24 1 23 } 5 2 1 17 y 5 24 1 7 }

3 2 1 17

y 5 2 7 } 5 y 5

23 } 3

The solutions are 1 23 } 5 , 2

7 } 5 2 and 1 7 }

3 ,

23 }

3 2 .


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33. y2 1 x 2 3 5 0 → x 5 3 2 y2

x2 2 4x 1 3y 1 1 5 0

(3 2 y2)2 2 4(3 2 y2) 1 3y 1 1 5 0

9 2 6y2 1 y4 2 12 1 4y2 1 3y 1 1 5 0

y4 2 2y2 1 3y 2 2 5 0

( y 2 1)( y3 1 y2 2 y 1 2) 5 0

(y 2 1)(y 1 2)( y2 2 y 1 1) 5 0

y 2 1 5 0 or y 1 2 5 0 or y2 2 y 1 1 5 0

y 5 1 or y 5 22 or No real solution

When y 5 1: When y 5 22:

12 1 x 2 3 5 0 (22)2 1 x 2 3 5 0

x 5 2 x 5 21


34. D;

x2 1 y2 1 6x 5 0 x2 1 y2 1 6x 5 0

y2 1 x 2 6 5 0 3 (21) 2y2 2 x 1 6 5 0

x2 1 5x 1 6 5 0

(x 1 3)(x 1 2) 5 0

x 5 23 or x 5 22


y2 1 (23) 2 6 5 0 y2 1 (22) 2 6 5 0

y2 5 9 y2 5 8

y 5 63 y 5 62 Ï}

2

There are four solutions.

35. When (1 2 y2)2 was expanded, the last term should have been y4, not y2. The correct substitution is:

(1 2 y2)2 1 y2 2 2(1 2 y2) 2 2y 5 21

1 2 2y2 1 y4 1 y2 2 2 1 2y2 2 2y 1 1 5 0

y4 1 y2 2 2y 5 0

y( y3 1 y 2 2) 5 0

y( y 2 1)( y2 1 y 1 2) 5 0

y 5 0 or y 2 1 5 0 or y2 1 y 1 2 5 0

y 5 1 No real solution


02 1 x 5 1 12 1 x 5 1

x 5 1 x 5 0


36. x2

} 2 1

y2

} 4 5 1 3 (216) 28x2 2 4y2 5 216

4y2 5 16 2 8x2 8x2 1 4y2 5 16

0 5 0

The equations are equivalent. There are infi nitely many solutions.

37. 3y2 1 x2 1 4x 1 18y 5 228

3y2 1 18y 1 (x2 1 4x 1 28) 5 0

y 5 218 6 Ï

}}}

182 2 4(3) 1 x2 1 4x 1 28 2 }}}

2(3)

y 5 218 6 Ï

}}

212x2 2 48x 2 12 }}} 6

y 5 29 6 Ï

}}

23(x2 1 4x 1 1) }} 3

9y2 2 4x2 1 8x 1 90y 5 2185

9y2 1 90y 1 (24x2 1 8x 1 185) 5 0

y 5 290 6 Ï

}}}

902 2 4(9) 1 24x2 1 8x 1 185 2 }}}

2(9)

y 5 290 6 Ï

}}

144x2 2 288x 1 1440 }}} 18

y 5 215 6 2 Ï

}}

x2 2 2x 1 10 }} 3


IntersectionX=-2.319489 Y=-2.017159

IntersectionX=-.2957947 Y=-2.821409

38. Solve the fi rst two equations:

x2 1 y2 5 1 3 (21)

x2 1 y2 1 4x 1 4y 2 5 5 0

2x2 2 y2 5 21

x2 1 y2 1 4x 1 4y 5 5

4x 1 4y 5 4

x 1 y 5 1

y 5 2x 1 1

When y 5 2x 1 1:

x2 1 (2x 1 1)2 5 1

x2 1 x2 2 2x 1 1 5 1

2x2 2 2x 5 0

2x (x 2 1) 5 0

x 5 0 or x 5 1

Substitute into third equation:


0 1 y 2 1 5 0 1 1 y 2 1 5 0

y 5 1 y 5 0



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583Algebra 2


Problem Solving

39. The distance the car travels is d 5 0.8t.

The police car travels a distance of d 5 2.5t2.

The police car catches up to the other car when the distances are equal.

0.8t 5 2.5t2

2.5t2 2 0.8t 5 0

t(2.5t 2 0.8) 5 0

t 5 0 or 2.5t 5 0.8

t 5 0.8

} 2.5 5 0.32

0.32 min 1 60 sec }

min 2 ø 19 seconds

The police car catches up to the other car in about 19 seconds.

40. y 5 2 x2

} 484 1 x 1 3

4y 2 x 5 2352, x ≥ 400

Solve the second equation for y.

4y 5 x 2 352

y 5 1 } 4 x 2 88

1 x } 4 2 88 2 1

x2

} 484 2 x 2 3 5 0

x2

} 488

2 3x

} 4 2 91 5 0

x2 2 366x 2 44,408 5 0

x 5 2(2366) 6 Ï

}}}

(2366)2 2 (4)(1)(244,408) }}}}

2(1)

x 5 366 6 Ï

}

311,588 }} 2

x ø 462, 296

Reject 296 because x ≥ 400.

When x 5 462:

y 5 462

} 4 2 88 5 27.5

The ball lands about 462 feet horizontally and 27.5 feet vertically from home plate.

41. a. A circle with a radius of 1 mile is represented by x2 1 y2 5 1.

Oak Lane has end points at (22, 1) and (5, 0).

m 5 0 2 1

} 5 2 (22)

5 2 1 } 7

Equation for Oak Lane:

y 2 0 5 2 1 } 7 (x 2 5)

y 5 2 1 } 7 x 1

5 } 7

b. Substitute y from Oak Lane in the equation for the circle.

x2 1 1 2 1 } 7 x 1

5 } 7 2 2 2 1 5 0

x2 1 1 } 49 x2 2

10 } 49 x 1

25 } 49 2 1 5 0

49x2 1 x2 2 10x 1 25 2 49 5 0

50x2 2 10x 2 24 5 0

(5x 2 4)(10x 1 6) 5 0

5x 2 4 5 0 or 10x 1 6 5 0

x 5 4 } 5 or x 5 2

3 } 5

When x 5 4 } 5 :

y 5 2 1 } 7 1 4 } 5 2 1

5 } 7 5 2

4 } 35 1

25 } 35 5

21 } 35 5

3 } 5

When x 5 2 3 } 5 :

y 5 2 1 } 7 1 2

3 } 5 2 1

5 } 7 5

3 } 35 1

25 } 35 5

28 } 35 5

4 } 5

The intersections of Oak Lane and the circle are at

1 2 3 } 5 ,

4 } 5 2 and 1 4 } 5 ,

3 } 5 2 .

c. Distance between the two points:

d 5 Î}}

1 4 } 5 2 3 } 5 2 2 1 1 3 } 5 2

4 } 5 2 2

5 Î}}

1 7 } 5 2 2 1 1 2 1 } 5 2 2

5 Ï}

49

} 25

1 1 } 25 5 Ï

}

2 ø 1.41

Students are not eligible for a parking pass on about 1.41 miles of Oak Lane.

42. x2 1 y2 1 60y 2 3456 5 0

x2 1 y 5 36 → x2 5 36 2 y

When x2 5 36 2 y:

(36 2 y) 1 y2 1 60y 2 3456 5 0

y2 1 59y 2 3420 5 0

( y 1 95)( y 2 36) 5 0

y 5 295 or y 5 36

When y 5 295: When y 5 36:

x2 5 36 2 (295) x2 5 36 2 36

x2 5 131 x2 5 0

x 5 6 Ï}

131 x 5 0

The solutions are (6 Ï}

131 , 295) and (0, 36).

Solve by elimination:

x2 1 y2 1 60y 2 3456 5 0

2x2 2 y 1 36 5 0

y2 1 59y 2 3420 5 0

This results in the same solutions found by using

substitution. Reject the solutions (6 Ï}

131 , 295) because a negative height does not make sense in this

situation. The ball will hit the ceiling directly above the net at 36 feet high.


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43. a. x2 2 y2 2 8x 1 8 5 0

2x2 1 y2 2 8y 1 8 5 0

28x 2 8y 1 16 5 0

28x 5 8y 2 16

x 5 2y 1 2

When x 5 2y 1 2:

y2 2 (2y 1 2)2 2 8y 1 8 5 0

y2 2 (y2 2 4y 1 4) 2 8y 1 8 5 0

24y 1 4 5 0

24y 5 24

y 5 1

When y 5 1: x 5 2(1) 1 2 5 1

The ship is located at (1, 1).

b. xy 2 24 5 0 → y 5 24

} x

x2 2 25y2 1 100 5 0

When y 5 24

} x :

x2 2 25 1 24 }

x 2 2 1 100 5 0

x2 2 14,400

} x2 1 100 5 0

x4 1 100x2 2 14,400 5 0

(x2 1 180)(x2 2 80) 5 0

x2 1 180 5 0 or x2 2 80 5 0

x2 5 2180 or x2 5 80

No real solution or x 5 6 Ï}

80 ø 68.94

Choose x ø 8.94 because the ship is west of the y-axis.

When x 5 28.94:

y 5 24

} x

y 5 24 }

(28.94) ø 22.68

The ship is located at (28.94, 22.68).

44. a. Lajitas: (0, 0)

Mexico City: (300, 2700)

La Paz: (2400, 2400)

The circle centered at Lajitas is x2 1 y2 5 7002.

The circle centered at Mexico City is (x 2 300)2 1 ( y 1 700)2 5 3002.

The circle centered at La Paz is (x 1 400)2 1 ( y 1 400)2 5 5002.

b. Lajitas and Mexico City:

x2 1 y2 5 7002 3 (21)

(x 2 300)2 1 (y 1 700)2 5 3002

2x2 2 y2 1 490,000 5 0

x2 1 y2 2 600x 1 1400y 1 490,000 5 0

2600x 1 1400y 1 980,000 5 0

23x 1 7y 1 4900 5 0

Lajitas and La Paz:

x2 1 y2 5 7002 3 (21)

(x 1 400)2 1 (y 1 400)2 5 5002

2x2 2 y2 1 490,000 5 0

x2 1 y2 1 800x 1 800y 1 70,000 5 0

800x 1 800y 1 560,000 5 0

x 1 y 1 700 5 0

New system of equations:

23x 1 7y 1 4900 5 0

x 1 y 1 700 5 0

c. When y 5 2x 2 700:

23x 1 7(2x 2 700) 1 4900 5 0

23x 2 7x 2 4900 1 4900 5 0

210x 5 0

x 5 0

When x 5 0

0 1 y 1 700 5 0

y 5 2 700

The epicenter is located 700 miles directly south of Lajitas.

d. Three stations are required to locate the epicenter because the intersection of three circles is a unique point.

45. Using the Pythagorean Theorem:

(10 2 x)2 1 (8 2 y)2 5 122

100 2 20x 1 x2 1 64 2 16y 1 y2 5 144

x2 1 y2 2 20x 2 16y 1 20 5 0

Because nABC , nCDE:

x }

8 2 y 5

y } 10 2 x

8y 2 y2 5 10x 2 x2

x2 2 y2 2 10x 1 8y 5 0

A system of equations is:

x2 2 y2 2 10x 1 8y 5 0 → y 5 4 6 Ï}}

x2 2 10x 1 16

The intersect feature on a graphing calculator shows that these graphs intersect at four points: (21.81, 10.12), (13.15, 23.58), (18.15, 16.81), and (0.51, 0.66). Reject the fi rst two solutions because x and y must be positive. Reject (18.15, 16.81) because these values exceed the dimensions of the mailbox. The solution is (0.51, 0.66).

Using the Pythagorean Theorem:

w2 5 x2 1 y2 5 0.512 1 0.662 ø 0.70

w ø 0.83

The width of the thickest box that will fi t is about 0.83 inch.


x2 1 y2 2 20x 2 16y 1 20 5 0 → y 5 8 6 Ï}}

2x2 1 20x 1 44

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585Algebra 2


Mixed Review

46. 25 5 (2)(2)(2)(2)(2) 5 32 47. 52 5 (5)(5) 5 25

48. 33 5 (3)(3)(3) 5 27

49. 46 5 (4)(4)(4)(4)(4)(4) 5 4096

50. 64 5 (6)(6)(6)(6) 5 1296 51. 162 5 (16)(16) 5 256

52. 113 5 (11)(11)(11) 5 1331

53. 95 5 (9)(9)(9)(9)(9) 5 59,049

54. 24, 4, 1

f (x) 5 (x 1 4)(x 2 4)(x 2 1)

5 (x 1 4)(x2 2 5x 1 4) 5 x3 2 5x2 1 4x 1 4x2 2 20x 1 16

5 x3 2 x2 2 16x 1 16

55. 23, 23, 0, 2

f (x) 5 (x 1 3)(x 1 3)(x)(x 2 2)

5 (x 1 3)(x 1 3)(x2 2 2x) 5 (x 1 3)(x3 2 2x2 1 3x2 2 6x) 5 (x 1 3)(x3 1 x2 2 6x) 5 x4 1 x3 2 6x2 1 3x3 1 3x2 2 18x

5 x4 1 4x3 2 3x2 2 18x

56. 2i, 22i

f (x) 5 (x 2 2i)(x 1 2i)

5 x2 1 2ix 2 2ix 2 4i2

5 x2 2 4i2

5 x2 1 4

57. 3, 2 Ï}

5

f (x) 5 (x 2 3)(x 1 Ï}

5 )(x 2 Ï}

5 ) 5 (x 2 3)(x2 2 5) 5 x3 2 3x2 2 5x 1 15

58. 21, 2, 23i

f (x) 5 (x 1 1)(x 2 2)(x 1 3i)(x 2 3i)

5 (x 1 1)(x 2 2)(x2 1 9) 5 (x 1 1)(x3 1 9x 2 2x2 2 18) 5 x4 1 9x2 2 2x3 2 18x 1 x3 1 9x 2 2x2 2 18

5 x4 2 x3 1 7x2 2 9x 2 18

59. 22, 0, 4, Ï}

10

f (x) 5 (x 1 2)(x)(x 2 4)(x 2 Ï}

10 )(x 1 Ï}

10 ) 5 (x 1 2)(x2 2 4x)(x2 2 10) 5 (x 1 2)(x4 2 10x2 2 4x3 1 40x) 5 x5 2 10x3 2 4x4 1 40x2 1 2x4 2 20x2 2 8x3 1 80x

5 x5 2 2x4 2 18x3 1 20x2 1 80x

60. Location of behive: (0, 0)

Location of heather patch: (27, 5)

Range of bees:

x2 1 y2 ≤ 92

(27)2 1 (5)2 ≤? 81

49 1 25 ≤? 81

74 ≤ 81 ✓

Yes, the patch is located within the bees’ range.

Quiz 9.6–9.7 (p. 664)

1. Ellipse

Vertices: (3, 210), (3, 6)

Foci: (3, 27), (3, 3)


1 3 1 3 }

2 ,

210 1 6 }

2 2 5 (3, 22)

h 5 3, k 5 22



c2 5 a2 2 b2

25 5 64 2 b2

b2 5 64 2 25 5 39

The major axis is vertical.

An equation is (x 2 3)2

} 39

1 ( y 1 2)2

} 64 5 1.

2. Parabola

Vertex: (25, 2)

Focus: (25, 21)

Because p < 0 the parabola opens down.

h 5 25, k 5 2

p 5 21 2 (2) 5 23

(x 2 (25))2 5 4(23)( y 2 2)

(x 1 5)2 5 212( y 2 2)

An equation is (x 1 5)2 5 212( y 2 2).

3. Hyperbola

Foci (23, 1), (6, 1)

Vertices: (0, 1), (3, 1)


1 0 1 3 }

2 ,

1 1 1 }

2 2 5 1 3 } 2 , 1 2

h 5 3 } 2 , k 5 1

The vertices are 3 }

2 units from the center, so a 5

3 } 2 and

a2 5 9 } 4 . The foci are

9 }

2 units from the center, so c 5

9 } 2

and c2 5 81

} 4 .

c2 5 a2 1 b2

81

} 4 5

9 } 4 1 b2

b2 5 81

} 4 2 9 } 4 5

72 } 4 5 18


An equation is:

1 x 2

3 } 2 2 2 }

9 }

4 2

( y 2 1)2

} 18 5 1


n2ws-09b.indd 585 6/27/06 11:15:48 AM

4. 9x2 2 4y2 2 36x 2 32y 2 64 5 0

A 5 9, B 5 0, C 5 24

B2 2 4AC 5 0 2 4(9)(24) 5 144


9x2 2 4y2 2 36x 2 32y 2 64 5 0

9(x2 2 4x) 2 4(y2 1 8y) 5 64

9(x2 2 4x 1 4) 2 4( y2 1 8y 1 16) 5 64 1 36 2 64

9(x 2 2)2 2 4(y 1 4)2 5 36

(x 2 2)2

} 4 2

( y 1 4)2

} 9 5 1

Center: (2, 24)

a2 5 4 a 5 2

b2 5 9 b 5 3

Vertices: (0, 24), (4, 24)

1

y

22 x

(2, 24)

(4, 24)

(2, 27)

(2, 21)

(0, 24)

5. 2x2 2 y2 2 4x 1 12y 1 129 5 0

A 5 21, B 5 0, C 5 21

B2 2 4AC 5 0 2 4(21)(21) 5 24

The conic is a circle because B2 2 4AC < 0,

B 5 0, and A 5 C.

2x2 2 y2 2 4x 1 12y 1 129 5 0

x2 1 4x 1 y2 2 12y 5 129

(x2 1 4x 1 4) 1 ( y2 2 12y 1 36) 5 129 1 4 1 36

(x 1 2)2 1 ( y 2 6)2 5 169

Center: (22, 6)

r 5 Ï}

169 5 13

4

y

24 x

(22, 6)

6. x2 1 6x 2 y 1 16 5 0

A 5 1, B 5 0, C 5 0

B2 2 4AC 5 0 2 4(1)(0) 5 0


x2 1 6x 2 y 1 16 5 0

x2 1 6x 5 y 2 16

x2 1 6x 1 9 5 y 2 16 1 9

(x 1 3)2 5 y 2 7

h 5 23, k 5 7

Vertex: (23, 7)

4p 5 1

p 5 1 } 4 5 0.25

The parabola opens up. The focus is 0.25 units above the vertex at (23, 7.25).

x

y

2

2

(23, 7)

(23, 7.25)

7. 12x2 1 45y2 1 120x 1 90y 2 150 5 0

A 5 12, B 5 0, C 5 45

B2 2 4AC 5 0 2 4(12)(45) 5 22160.


12x2 1 45y2 1 120x 1 90y 2 150 5 0

12(x2 1 10x) 1 45(y2 1 2y) 5 150

12(x2 1 10x 1 25) 1 45(y2 1 2y 1 1) 5 150 1 300 1 45

12(x 1 5)2 1 45(y 1 1)2 5 495

(x 1 5)2

} 41.25

1 ( y 1 1)2

} 11 5 1

h 5 25, k 5 21

Center: (25, 21)

a2 5 41.25 a 5 6.42

b2 5 11 b 5 3.31

Vertices: (211.42, 21) and (1.42, 21)

Co-vertices: (25, 2.31) and (25, 24.31)

x

y4

4

(25, 2.31)

(25, 21)

(25, 24.31)

(1.42, 21)

(211.42, 21)

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n2ws-09b.indd 586 6/27/06 11:15:55 AM

8. x 1 2y2 5 26

x 1 8y 5 0 → x 5 28y

28y 1 2y2 5 26

2y2 2 8y 1 6 5 0

2(y2 2 4y 1 3) 5 0

2(y 2 3)(y 2 1) 5 0

y 5 1 or y 5 3


x 5 28(1) 5 28 x 5 28(3) 5 224


9. x2 1 4x 1 y2 1 6y 2 12 5 0

2x 2 y 5 4 → y 5 2x 2 4

x2 1 4x 1 (2x 2 4)2 1 6(2x 2 4) 2 12 5 0

x2 1 4x 1 (4x2 2 16x 1 16) 1 12x 2 24 2 12 5 0

5x2 2 20 5 0

5x2 5 20

x2 5 4

x 5 62


2(2) 2 y 5 4 2(22) 2 y 5 4

2y 5 0 2y 5 8

y 5 0 y 5 28


10. x2 2 y 2 4 5 0 3 (21) 2x2 1 y 1 4 5 0

x2 1 3y2 2 4y 2 10 5 0 x2 1 3y2 2 4y 2 10 5 0

3y2 2 3y 2 6 5 0

3( y2 2 y 2 2) 5 0

3( y 2 2)( y 1 1) 5 0

y 5 2 or y 5 21


x2 2 2 2 4 5 0 x2 2 (21) 2 4 5 0

x2 5 6 x2 5 3

x 5 6 Ï}

6 x 5 6 Ï}

3

The solutions are (6 Ï}

6 , 2) and (6 Ï}

3 , 21). 11. y2 2 6x 2 2y 2 3 5 0 3 (22) 22y2 1 12x 1 4y 1 6 5 0

2y2 2 4y 1 x 1 6 5 0 2y2 1 x 2 4y 1 6 5 0

13x 1 12 5 0

x 5 2 12

} 13

When x 5 2 12

} 13 :

y2 2 6 1 2 12

} 13 2 2 2y 2 3 5 0

y2 2 2y 1 33

} 13 5 0

y 5 2(22) 6 Î}}

(22)2 2 4(1) 1 33 }

13 2 }}}

2(1)

y 5 4 6 Î}

280 }

13 } 2


12. y2 2 4x2 2 4y 5 0 3 (21)

2x2 1 y2 2 8x 2 4y 5 28

4x2 2 y2 1 4y 5 0

2x2 1 y2 2 8x 2 4y 5 28

6x2 2 8x 5 28

2(3x2 2 4x 1 4) 5 0

x 5 2(24) 6 Ï

}}

(24)2 2 4(3)(4) }}}

2(3)

x 5 4 6 Ï

}

232 } 6


13. 16x2 1 9y2 1 32x 2 18y 5 119

x2 1 y2 1 2x 1 6y 5 15 3 (216)

16x2 1 9y2 1 32x 2 18y 5 119

216x2 2 16y2 2 32x 2 96y 5 2240

27y2 2 114y 5 2121

(7y 1 121)( y 2 1) 5 0

y 5 2 121

} 7 or y 5 1

When y 5 2 121

} 7 :

x2 1 1 2 121

} 7 2 2 1 2x 1 6 1 2 121

} 7 2 5 15

x2 1 2x 1 8824

} 49 5 0

x 5 22 6 Ï

}}

22 2 4(1) 1 8824 }

49 2 }}

2(1)

x 5 22 6 Î}

235,100 }

49 }} 2

No real solution

When y 5 1:

x2 1 12 1 2x 1 6(1) 5 15

x2 1 2x 2 8 5 0

(x 1 4)(x 2 2) 5 0

x 5 24, 2


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587Algebra 2



n2ws-09b.indd 587 6/27/06 11:16:00 AM

14. First station: x2 1 y2 5 102

Second station: (x 2 20)2 1 (y 2 15)2 5 152

x2 2 40x 1 400 1 y2 2 30y 1 225 5 225

x2 1 y2 2 40x 2 30y 5 2400

System of equations:

x2 1 y2 5 100

x2 1 y2 2 40x 2 30y 5 2400 3 (21)

x2 1 y2 5 100

2x2 2 y2 1 40x 1 30y 5 400

40x 1 30y 5 500

y 5 2 4 } 3 x 1

50 } 3

x2 1 1 2 4 } 3 x 1

50 } 3 2 2 5 100

x2 1 16

} 9 x2 2 400

} 9 x 1 2500

} 9 5 100

25

} 9 x2 2

400 } 9 x 1

1600 } 9 5 0

25

} 9 (x2 2 16x 1 64) 5 0

25

} 9 (x 2 8)2 5 0

x 5 8

When x 5 8: y 5 2 4 } 3 (8) 1

50 } 3 5 6

The ship is 8 miles east and 6 miles north of the fi rst station. Only one location is possible because the circles intersect at only one point.

9.7 Extension (p. 666)

1. 7(x 2 3)2 1 7( y 1 7)2 5 56

(x 2 3)2 1 ( y 1 7)2 5 8

The conic is a circle, so its eccentricity is e 5 0.

2. 16(x 1 1)2 2 9(y 2 5)2 5 144

(x 1 1)2

} 9 2

(y 2 5)2

} 16 5 1

The conic is a hyperbola.

a2 5 9 a 5 3

b2 5 16 b 5 4

c 5 Ï}

a2 1 b2 5 Ï}

25 5 5

e 5 c } a 5

5 } 3 ø 1.67

3. (x 2 6)2

} 49

1 (y 2 5)2

} 64 5 1

The conic is an ellipse.

a2 5 64 a 5 8

b2 5 49 b 5 7

c 5 Ï}

a2 2 b2 5 Ï}

15

e 5 c } a 5

Ï}

15 } 8 ø 0.48

4. ( y 2 4)2

} 100

2 (x 1 2)2

} 9 5 1


a2 5 100 a 5 10

b2 5 9 b 5 3

c 5 Ï}

a2 1 b2 5 Ï}

109

e 5 c } a 5

Ï}

109 } 10 ø 1.04

5. (x 2 5)2 5 10y

The conic is a parabola, so e 5 1

6. 81(x 1 4)2 1 ( y 2 9)2 5 81

(x 1 4)2 1 ( y 2 9)2

} 81 5 1


a2 5 81, a 5 9

b2 5 1, b 5 1

c 5 Ï}

a2 2 b2 5 Ï}

80 5 4 Ï}

5

e 5 c } a 5

4 Ï}

5 } 9 ø 0.99

7. Ellipse; vertices (26, 4), (6, 4); e 5 0.4

Center: 1 26 1 6 }

2 ,

4 1 4 }

2 2 5 (0, 4)

The vertices are 6 units from the center, so a 5 6.

e 5 c } a

4 }

10 5

c } 6

c 5 12

} 5

c2 5 a2 2 b2

1 12 } 5 2 2 5 62 2 b2

b2 5 36 2 144

} 25 5 756

} 25

An equation for the ellipse is x2

} 36

1 25( y 2 4)2

} 756 5 1.

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n2ws-09b.indd 588 6/27/06 11:16:04 AM

8. Ellipse; foci (24, 2), (24, 22); e 5 0.5

Center: 1 24 1 (24) }

2 ,

2 1 (22) }

2 2 5 (24, 0)

The foci are 2 units from the center, so c 5 2.

e 5 c } a

5 }

10 5

2 } a

a 5 4

c2 5 a2 2 b2

4 5 16 2 b2

b2 5 16 2 4 5 12

An equation for the ellipse is (x 1 4)2

} 12

1 y2

} 16 5 1.

9. Ellipse; center (0, 5); vertex (7, 5); e 5 0.2

The vertex is 7 units from the center, so a 5 7.

e 5 c } a

2 }

10 5

c } 7

c 5 14

} 10 5 7

} 5

c2 5 a2 2 b2

1 7 } 5 2 2 5 49 2 b2

b2 5 49 2 49

} 25 5 1176

} 25

An equation for the ellipse is x2

} 49

1 25(y 2 5)2

} 1176 5 1.

10. Hyperbola; foci (4, 25), (4, 3); e 5 2.5 5 5 } 2

Center: 1 4 1 4 }

2 ,

25 1 3 }

2 2 5 (4, 21)

The foci are 4 units from the center, so c 5 4.

e 5 c } a

5 }

2 5

4 } a

a 5 8 } 5

c2 5 a2 1 b2

42 5 1 8 } 5 2 2 1 b2

b2 5 16 2 64

} 25 5 336

} 25

The hyperbola’s equation is 25( y 1 1)2

} 64

2 25(x 2 4)2

} 336 5 1.

11. Hyperbola; vertices (1, 24), (7, 24); e 5 1.8 5 18

} 10 5 9 } 5

Center: 1 1 1 7 }

2 ,

24 1 (24) }

2 2 5 (4, 24)

The vertices are 3 units from the center, so a 5 3.

e 5 c } a

9 } 5 5

c } 3

c 5 27

} 5

c2 5 a2 1 b2

1 27 } 5 2 2 5 9 1 b2

b2 5 729

} 25 2 9 5 504

} 25

The hyperbola’s equation is (x 2 4)2

} 9 2

25( y 1 4)2

} 504 5 1.

12. Hyperbola; center (22, 3); focus (25, 3); e 5 4

The focus is 3 units from the center, so c 5 3.

e 5 c } a

4 5 3 } a

a 5 3 } 4

c2 5 a2 1 b2

9 5 1 3 } 4 2 2 1 b2

b2 5 9 2 9 } 16 5

135 } 16

The hyperbola’s equation is 16(x 1 2)2

} 9 2

16( y 2 3)2

} 135 5 1.

13. 2a 5 11

a 5 11

} 2 5 5.5

e 5 c } a

0.751 5 c } 5.5

c 5 (0.751)(5.5) ø 4.13

c2 5 a2 2 b2

(4.13)2 5 (5.5)2 2 b2

b2 ø 13.19

An equation of Nereid’s orbit is x2 }

30.25 1

y2

} 13.19 5 1.

14.

x

y

Earth

14,300

The distance from the closest point is a 2 c 5 14,300. So, c 5 a 2 14,300.

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589Algebra 2



n2ws-09b.indd 589 6/27/06 11:16:09 AM

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e 5 c } a

0.394 5 a 2 14,300

} a

0.394 5 1 2 14,300

} a

20.606 5 2 14,300

} a

a ø 23,597

The length of the major axis is about 2a 5 47,194 kilometers.

The satellites distance from Earth at its farthest point in the orbit is about 47,194 2 14,300 5 32,894 kilometers.

15. In the ellipse, 0 < c < a, therefore 0 < c }

a < 1, and

0 < e < 1.

In the hyperbola, 0 < a < c, therefore c }

a > 1, and e > 1.

Mixed Review of Problem Solving (p. 667)

1. a. The fi rst dish is at the origin with vertex (0, 0). Its focus is at (1.5, 0), so p 5 1.5.

The equation is y2 5 4(1.5)x 5 6x.

The second dish is located with its vertex at (47, 0). h 5 47, k 5 0.

It is facing the other direction, so p 5 21.5. Its equation is (y 2 0)2 5 26(x 2 47), or y2 5 26(x 2 47).

b. Radius 5 d } 2 5

67 in. } 2 p 1 ft.

} 12 in.

ø 2.79 feet

When y 5 2.79:

2.792 5 6x

1.3 ø x Each dish is about 1.3 feet deep. c.

x

y(1.25, 4)

focus

3 ft15

8 ft

y2 5 4px

(4)2 5 4p(1.25)

12.8 5 4p

p 5 3.2

y2 5 12.8x

The vertex is at (0, 0) and the focus is at (3.2, 0). The distance between them is 3.2 feet.

2.

x

y

4

4

(8, 0) (12, 0)

(28, 0)

(212, 0)

a. a 5 8, a2 5 64

c 5 12, c2 5 144

c2 5 a2 1 b2

144 5 a2 1 b2

b2 5 144 2 64 5 80

An equation for the mirror is x2

} 64

2 y2

} 80 5 1.

b. The light beam originates at (0, 9) and travels toward the focus, (12, 0).

m 5 9 2 0

} 0 2 12 5 2 3 } 4

An equation for the path of the beam is

y 5 2 3 } 4 (x 2 12) or y 5 2

3 } 4 x 1 9.

c. The intersection is the solution of the system:

y 5 2 3 } 4 x 1 9

x2

} 64

2 y2

} 80 5 1 → 5x2 2 4y2 5 320

When y 5 2 3 } 4 x 1 9:

5x2 2 4 1 2 3 } 4 x 1 9 2 2 5 320

5x2 2 4 1 9 } 16

x2 2 54

} 4 x 1 81 2 2 320 5 0

5x2 2 9 } 4 x2 1 54x 2 324 2 320 5 0

11

} 4 x2 1 54x 2 644 5 0

11x2 1 216x 2 2576 5 0

x 5 2216 6 Ï

}}

2162 2 4(11)(22576) }}}

2(11)

x ø 2216 6 400 }

22

x ø 8.36 or x 5 228

x > 0, so x 5 8.36

y 5 2 3 } 4 (8.36) 1 9 5 2.73

The light beam reflects off the mirror at about (8.36, 2.73).

3. Sample answer:


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591Algebra 2


4. a. x2

} 36

2 y2

} 100 5 1

The vertices (66, 0) are the closest points to the origin. So, the shortest possible horizontal distance you could be from the jet when you fi rst hear the sonic boom is 6 miles away.

b. x2

} 9 2

y2

} 25 5 1

The second time the jet passes, the sound will be heard at a horizontal distance of Ï

}

9 5 3 miles.

c. The hyperbola in part (b) is 1 }

4 the hyperbola in part (c).

5. a. Mirror A:

y2 2 72x 2 450 5 0

y2 5 72x 1 450

The mirror is in the form of a parabola.

b. Mirror B:

88.4x2 2 49.7y2 5 4390

88.4x2

} 4390

2 49.7y2

} 4390 5 1

x2 }

49.7 2

y2

} 88.3 5 1

The mirror is in the form of a hyperbola.

6. 9x2 1 y2 1 8y 5 20

x2 1 4y2 5 16 3 (29)

9x2 1 y2 1 8y 2 20 5 0

29x2 2 36y2 1 144 5 0

235y2 1 8y 1 124 5 0

2(35y2 2 8y 2 124) 5 0

2(35y 1 62)( y 2 2) 5 0

35y 5 262 or y 5 2

y 5 2 62

} 35

The y-coordinate of the uppermost point of intersection of the ellipses is 2.

Chapter 9 Review (pp. 669–672)

1. A parabola is the set of all points in a plane equidistant from a point called the focus and a line called the directrix.

2. The line segment joining the two co-vertices of an ellipse is the minor axis.

3. The line segment joining the two vertices of a hyperbola is the transverse axis.

4. The asymptotes of a hyperbola help in drawing the graph by indicating how wide or narrow the hyperbola is.

5. (26, 25), (2, 23)

d 5 Ï}}}

(26 2 2)2 1 (25 2 (23))2

5 Ï}}

(28)2 1 (22))2 5 Ï}

64 1 4 5 Ï}

68 5 2 Ï}

17

M 1 26 1 2 }

2 ,

25 1 (23) }

2 2 5 1 24

} 2 ,

28 }

2 2 5 (22, 24)

6. (22, 5), (1, 9)

d 5 Ï}}

(22 2 1)2 1 (5 2 9)2

5 Ï}}

(23)2 1 (24)2 5 Ï}

9 1 16 5 Ï}

25 5 5

M 1 22 1 1 } 2 ,

5 1 9 }

2 2 5 1 2

1 } 2 , 7 2

7. (23, 24) , (2, 5)

d 5 Ï}}

(23 2 2)2 1 (24 2 5)2

5 Ï}}

(25)2 1 (29)2 5 Ï}

25 1 81 5 Ï}

106

M 1 23 1 2 } 2 ,

24 1 5 }

2 2 5 1 2

1 } 2 ,

1 }

2 2

8. (2200, 40) , (30, 2140)

d 5 Ï}}}

(2200 2 30)2 1 (40 2 (2140))2

5 Ï}}

(2230)2 1 (180)2 5 Ï}}

52,900 1 32,400

5 Ï}

85,300 ø 292

The distance between the skydivers is about 292 yards.

9. x2 5 16y

x

y

2

2

(0, 4)

y 5 24

4p 5 16

p 5 4

The focus is (0, 4).

The directrix is y 5 24.

The axis of symmetry is vertical, x 5 0.

10. y2 5 26x

32

( )32 , 02

x 5

3

3

y

x

4p 5 26

p 5 2 3 } 2

The focus is 1 2 3 } 2 , 0 2 .

The directrix is x 5 3 } 2 .

The axis of symmetry is horizontal, y 5 0.

11. x2 1 4y 5 0

x

y

2

3

y 5 1

(0, 21)

x2 5 24y

4p 5 24

p 5 21


The directrix is y 5 1.

The axis of symmetry is vertical, x 5 0.

12. Focus: (25, 0)

p 5 25

The equation is y2 5 4(25)x, or y2 5 220x.

13. Focus: (0, 3)

p 5 3

The form of the equation is x2 5 4py.

The equation is x2 5 12y.


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14. Directrix: x 5 26

p 5 6

The form of the equation is y2 5 4px.

The equation is y2 5 24x.

15. x2 1 y2 5 81

2

x

y

22

(0, 29)

(0, 9)

(9, 0)(29, 0)

r2 5 81

r 5 9

16. x2 5 40 2 y2

2

y

22 x

10, 02

10, 022

( )100, 2( )

100, 22( )( )

x2 1 y2 5 40

r2 5 40

r 5 Ï}

40 5 2 Ï}

10

17. 3x2 1 3y2 5 147 (0, 7)

(0, 27)

(7, 0)(27, 0)2

2 x

y

x2 1 y2 5 49

r2 5 49

r 5 7

18. (5, 9)

(5)2 1 (9)2 5 106

The standard form of the equation is x2 1 y2 5 106.

19. (28, 2)

(28)2 1 (2)2 5 68


20. (27, 24)

(27)2 1 (24)2 5 65


21. 16x2 1 25y2 5 400

1

x

y

21

(3, 0)(23, 0)

(5, 0)(25, 0)

(0, 4)

(0, 24)

x2

} 25

1 y2

} 16 5 1

a2 5 25 a 5 5

b2 5 16 b 5 4

Vertices: (5, 0) and (25, 0)


c2 5 a2 2 b2 5 25 2 16 5 9

c 5 3

Foci: (3, 0) and (23, 0)

22. 81x2 1 9y2 5 729

2

y

24 x

(0, 9)

(0, 29)

(3, 0)(23, 0)

20, 6

0, 26 2

( )

( )

x2

} 9 1

y2

} 81 5 1

a2 5 81 a 5 9

b2 5 9 b 5 3

c2 5 a2 2 b2 5 81 2 9 5 72

c 5 Ï}

72 5 6 Ï}

2

Vertices: (0, 69)


Foci: 1 0, 66 Ï}

2 2 23. 64x2 1 36y2 5 2304

2

2 x

y(0, 8)

(0, 28)

(6, 0)(26, 0)

70, 2

70, 22

( )

( )

x2

} 36

1 y2

} 64 5 1

a2 5 64 a 5 8

b2 5 36 b 5 6

c2 5 64 2 36 5 28

c 5 Ï}

28 5 2 Ï}

7

Vertices: (0, 68)


Foci: (0, 62 Ï}

7 ) 24. Vertex: (26, 0)

Co-vertex: (0, 23)

a 5 6 a2 5 36

b 5 3 b2 5 9

An equation is x2

} 36

1 y2

} 9 5 1.

25. Vertex: (0, 28)

Focus: (0, 5)

a 5 8 a2 5 64

c 5 5 c2 5 25

c2 5 a2 2 b2

25 5 64 2 b2

b2 5 64 2 25 5 39

An equation is x2

} 39

1 y2

} 64 5 1.

26. 9x2 2 y2 5 9 4

y

22 x

10, 0( )10, 0( )2

(1, 0)(21, 0)

(0, 3)

(0, 23)

x2 2 y2

} 9 5 1

a2 5 1 a 5 1

b2 5 9 b 5 3

c2 5 a2 1 b2 5 1 1 9 5 10

c 5 Ï}

10

Vertices: (61, 0)

Foci: (6 Ï}

10 , 0) Asymptotes: y 5 63x.


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593Algebra 2


27. 4x2 2 16y2 5 64 4

x

y

26

(0, 2)

(0, 22)

(4, 0)(24, 0)

5, 02

5, 022( )

( )

x2

} 16

2 y2

} 4 5 1

a2 5 16 a 5 4

b2 5 4 b 5 2

c2 5 16 1 4 5 20

c 5 Ï}

20 5 2 Ï}

5

Vertices: (64, 0)

Foci: (62 Ï}

5 , 0)


28. 100y2 2 36x2 5 3600

8

y

212 x

(0, 6)

(0, 26)

(10, 0)(210, 0)

340, 2

3422

( )

( )0,

y2

} 36

2 x2

} 100 5 1

a2 5 36 a 5 6

b2 5 100 b 5 10

c2 5 36 1 100 5 136

c 5 Ï}

136 5 2 Ï}

34

Vertices: (0, 66)

Foci: (0, 62 Ï}

34 )


29. Foci: (0, 65)

Vertices: (0, 62)

a 5 2 a2 5 4

c 5 5 c2 5 25

c2 5 a2 1 b2

25 5 4 1 b2

b2 5 25 2 4 5 21

An equation is y2

} 4 2

x2

} 21 5 1.

30. Foci: (69, 0)

Vertices: (64, 0)

a 5 4 a2 5 16

c 5 9 c2 5 81

c2 5 a2 1 b2

81 5 16 1 b2

b2 5 81 2 16 5 65

An equation is x2

} 16

2 y2

} 65 5 1.

31. 4x2 1 9y2 1 40x 1 72y 1 208 5 0

A 5 4, B 5 0, C 5 9

B2 2 4AC 5 0 2 4(4)(9) 5 2144


4x2 1 9y2 1 40x 1 72y 1 208 5 0

4(x2 1 10x) 1 9(y2 1 8y) 5 2208

4(x2 1 10x 1 25) 1 9(y2 1 8y 1 16) 5 2208 1 100 1 144

4(x 1 5)2 1 9(y 1 4)2 5 36

(x 1 5)2

} 9 1

(y 1 4)2

} 4 5 1

h 5 210, k 5 24

(213, 24)

(27, 24)

(210, 22)(210, 24)

(210, 26)

2

x

y

22 a2 5 9, a 5 3

b2 5 4, b 5 2


The vertices are 3 units left and right of the center at (213, 24) and (27, 24).

The co-vertices are 2 units above and below the center at (210, 22) and (210, 26).

32. y2 2 10y 2 8x 1 1 5 0

A 5 0, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(0)(1) 5 0


y2 2 10y 2 8x 1 1 5 0

y2 2 10y 5 8x 2 1

y2 2 10y 1 25 5 8x 2 1 1 25

( y 2 5)2 5 8x 1 24

( y 2 5)2 5 8(x 1 3)

h 5 23, k 5 5

The vertex is at (23, 5).

4p 5 8

p 5 2

The paprabola opens to the right, so the focus is 2 units to the right of the vertex at (21, 5).

x

y

2

4

(23, 5)

(21, 5)

33. 9x2 2 y2 2 18x 2 4y 2 5 5 0

A 5 9, B 5 0, C 5 21

B2 2 4AC 5 0 2 4(9)(21) 5 36


9x2 2 y2 2 18x 2 4y 2 5 5 0

9(x2 2 2x) 2 (y2 1 4y) 5 5

9(x2 2 2x 1 1) 2 (y2 1 4y 1 4) 5 5 1 9 2 4

9(x 2 1)2 2 (y 1 2)2 5 10

(x 2 1)2

} 10

} 9 2

(y 1 2)2

} 10 5 1

h 5 1, k 5 22

The center is at (1, 22).


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a2 5 10

} 9 a 5 Ï

}

10 } 3

b2 5 10 b 5 Ï}

10

Vertices: 1 16 Ï

}

10 } 3 , 22 2

x

y

21

(1, 22)

1031 1 , 22( )10

31 2 , 22( )

1022 1

1022 2

( )

( )

34. x2 1 y2 1 4x 2 14y 1 17 5 0

A 5 1, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(1)(1) 5 24

The conic is a circle because B2 2 4AC < 0, b 5 0, andA 5 C.

x2 1 y2 1 4x 2 14y 1 17 5 0

(x2 1 4x) 1 ( y2 2 14y) 5 217

(x2 1 4x 1 4) 1 ( y2 2 14y 1 49) 5 217 1 4 1 49

(x 1 2)2 1 ( y 2 7)2 5 36

The radius is 6.

h 5 22, k 5 7


4

y

22 x

(22, 7)

35. y2 5 4x → x 5 1 } 4 y2

2x 2 5y 5 28

2 1 1 } 4 y2 2 2 5y 5 28

1 }

2 y2 2 5y 1 8 5 0

1 }

2 ( y2 2 10y 1 16) 5 0

1 }

2 ( y 2 8)(y 2 2) 5 0

y 5 8 or y 5 2

When y 5 8: x 5 1 } 4 (8)2 5 16

When y 5 2: x 5 1 } 4 (2)2 5 1


36. x2 1 y2 2 100 5 0

x 1 y 2 14 5 0 → x 5 14 2 y

(14 2 y)2 1 y2 2 100 5 0

196 2 28y 1 y2 1 y2 2 100 5 0

2y2 2 28y 1 96 5 0

2(y2 2 14y 1 48) 5 0

2(y 2 8)(y 2 6) 5 0

y 5 8 or y 5 6

When y 5 8: x 5 14 2 (8) 5 6

When y 5 6: x 5 14 2 (6) 5 8


37. 16x2 2 4y2 2 64 5 0

4x2 1 9y2 2 40x 1 64 5 0

36x2 2 9y2 2 144 5 0

4x2 1 9y2 2 40x 1 64 5 0

40x2 2 40x 2 80 5 0

40 1 x2 2 x 2 2 2 5 0

40(x 1 1)(x 2 2) 5 0

x 5 21 or x 5 2


16(21)2 2 4y2 2 64 5 0 16(2)2 2 4y2 2 64 5 0

24y2 5 48 24y2 5 0

y2 5 224 y 5 0

No real solution


Chapter 9 Test (p. 673)

1. (21, 5), (7, 3)

d 5 Ï}}

(21 2 7)2 1 (5 2 3)2

5 Ï}}

(28)2 1 (2)2

5 Ï}

64 1 4 5 Ï}

68 5 2 Ï}

17

M 1 21 1 7 } 2 ,

5 1 3 }

2 2 5 (3, 4)

2. (4, 2),(8, 8)

d 5 Ï}}

(4 2 8)2 1 (2 2 8)2

5 Ï}}

(24)2 1 (26)2

5 Ï}

16 1 36 5 Ï}

52 5 2 Ï}

13

M 1 4 1 8 } 2 ,

2 1 8 }

2 2 5 (6, 5)

3. (21, 26), (1, 5)

d 5 Ï}}

(21 2 1)2 1 (26 2 5)2

5 Ï}}

(22)2 1 (211)2

5 Ï}

4 1 121 5 Ï}

125 5 5 Ï}

5

M 1 21 1 1 } 2 ,

26 1 5 }

2 2 5 1 0, 2

1 } 2 2


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595Algebra 2


4. (2, 25), (3, 1)

d 5 Ï}}

(2 2 3)2 1 (25 2 1)2

5 Ï}}

(21)2 1 (26)2

5 Ï}

1 1 36 5 Ï}

37

M 1 2 1 3 } 2 ,

25 1 1 }

2 2 5 1 5 } 2 , 22 2

5. (26, 22), (23, 5)

d 5 Ï}}}

(26 2 (23))2 1 (22 2 5)2

5 Ï}}

(23)2 1 (27)2

5 Ï}

9 1 49 5 Ï}

58

M 1 26 1 (23) } 2 ,

22 1 5 }

2 2 5 1 2

9 } 2 ,

3 }

2 2

6. (1, 9), (10, 22)

d 5 Ï}}}

(1 2 10)2 1 (9 2 (22))2

5 Ï}}

(29)2 1 (11)2

5 Ï}

81 1 12 1 5 Ï}

202

M 1 1 1 10 } 2 ,

9 1 (22) }

2 2 5 1 11

} 2 , 7 }

2 2

7. y2 2 24x 5 0

y2 5 24x

The conic is a parabola.

4p 5 24

p 5 6


2

y

22 x

(0, 0) (6, 0)

8. x2 1 y2 5 16

The conic is a circle with radius r 5 Ï}

16 5 4.

(4, 0)(24, 0)

(0, 4)

(0, 24)

(0, 0)

1

x

y

21

9. 64y2 2 x2 5 64

y2 2 x2

} 64 5 1


a2 5 1 a 5 1

2y

29 x

(8, 0)(28, 0)

(0, 1)

(0, 21) b2 5 64 b 5 8

Vertices: (0, 61)

10. 18x2 1 2y2 5 18

x2 1 y2

} 9 5 1


1

x

y

22

(0, 3)

(0, 23)

(1, 0)(21, 0)

a2 5 9 a 5 3

b2 5 1 b 5 1

Vertices: (0, 63)


11. (x 2 6)2 1 ( y 1 1)2 5 36

The conic is a circle with radius r 5 Ï}

36 5 6 and center (6, 21).

2

x

y

22(6, 21)

12. (x 1 4)2 5 6(y 2 2)

h 5 24, k 5 2

The conic is a parabola with vertex (24, 2). The parabola opens up.

4p 5 6

p 5 3 } 2

Focus: 1 24, 2 1 3 } 2 2 5 1 24, 3

1 }

2 2

1

21 x

y

1224, 3( )

(24, 2)

13. (x 1 4)2

} 9 2

( y 2 7)2

} 49 5 1


h 5 24 k 5 7

Center: (24, 7)

a2 5 9 a 5 3

b2 5 49 b 5 7

The vertices are 3 units left and right of the center at (27, 7) and (21, 7).

y

2 x

(24, 14)

(24, 0)

(24, 7)

(21, 7)(27, 7)

24


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14. (x 2 8)2

} 81

1 ( y 2 2)2

} 100 5 1


h 5 8, k 5 2

Center: (8, 2)

a2 5 100 a 5 10

b2 5 81 b 5 9

The vertices are 10 units above and below the center at (8, 12) and (8, 28).

The co-vertices are 9 units left and right of the center at (17, 2) and (21, 2).

8

x

y

22

(8, 12)

(8, 2)

(8, 28)

(17, 2)

(21, 2)

15. ( y 2 5)2

} 9 2 (x 1 3)2 5 1

The conic is a hyperbola. y

x

2

1

(23, 8)

(23, 5)

(23, 2)

(24, 5)(22, 5)

h 5 23, k 5 5

Center: (23, 5)

a2 5 9 a 5 3

b2 5 1 b 5 1

The vertices are 3 units above and below the center at (23, 8) and (23, 2).

16. Parabola; vertex: (0, 0); directrix: x 5 26

The form of the equation is y2 5 4pxp 5 6

The equation is y2 5 24x.

17. Parabola; vertex: (22, 21); focus: (22, 5)

The form of the equation is (x 2 h)2 5 4p ( y 2 k).

h 5 22, k 5 21

p 5 5 2 (21) 5 6

The equation is (x 1 2)2 5 24( y 1 1).

18. Circle; center: (0, 0); passes through (25, 2)

x2 1 y2 5 r2

(25)2 1 (2)2 5 r2

29 5 r2

The equation is x2 1 y2 5 29.

19. Circle; center: (1, 24); radius 5 6

h 5 1, k 5 24

The equation is (x 2 1)2 1 ( y 1 4)2 5 36.

20. Ellipse; center: (0,0); vertex: (0, 6); co-vertex: (23, 0)

The form of the equation is x2

} b2 1

y2

} a2 5 1.

a 5 6, a2 5 36

b 5 3 b2 5 9

The equation is x2

} 9 1

y2

} 36 5 1.

21. Ellipse; vertices: (21, 4) and (7, 4); foci: (1, 4) and (5, 4)

The form of the equation is (x 2 h)2

} a2 1

( y 2 k)2

} b2 5 1.

The center is the midpoint of the vertices at

1 21 1 7 }

2 ,

4 1 4 }

2 2 5 (3, 4).

h 5 3, k 5 4



c2 5 a2 2 b2

4 5 16 2 b2

b2 5 16 2 4 5 12

The equation is (x 2 3)2

} 16

1 ( y 2 4)2

} 12 5 1.

22. Hyperbola; vertices: (0, 66); foci: (0, 69)

The form of the equation is y2

} a2 2

x2

} b2 5 1.

a 5 6 a2 5 36

c 5 9 c2 5 81

c2 5 a2 1 b2

b2 5 81 2 36 5 45

The equation is y2

} 36

2 x2

} 45 5 0.

23. Hyperbola: vertex: (2, 25); focus: (21, 25);

center: (5, 25)

The form of the equation is (x 2 h)2

} a2 2

( y 2 k)2

} b2 5 1.

h 5 5, k 5 25

The vertex is 3 units from the center, so a 5 3 and a2 5 9.

The focus is 6 units from the center, so c 5 6 and c2 5 36.

c2 5 a2 1 b2

36 5 9 1 b2

b2 5 36 2 9 5 27

The equation is (x 2 5)2

} 9 2

( y 1 5)2

} 27 5 1.


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597Algebra 2


24. x2 1 4y2 2 6x 2 16y 1 21 5 0

A 5 1, B 5 0, C 5 4

B2 2 4AC 5 0 2 4(1)(4) 5 216


x2 1 4y2 2 6x 2 16y 1 21 5 0

1 x2 2 6x 2 1 4(y2 2 4y) 5 221

(x2 2 6x 1 9) 1 4( y2 2 4y 1 4) 5 221 1 9 1 16

(x 2 3)2 1 4( y 2 2)2 5 4

(x 2 3)2

} 4 1 ( y 2 2)2 5 1

25. x2 1 y2 1 8x 1 12y 1 3 5 0

A 5 1, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(1)(1) 5 24

The conic is a circle because B2 2 4AC < 0, B 5 0, andA 5 C.

x2 1 y2 1 8x 1 12y 1 3 5 0

(x2 1 8x) 1 ( y2 1 12y) 5 23

(x2 1 8x 1 16) 1 ( y2 1 12y 1 36) 5 23 1 16 1 36

(x 1 4)2 1 ( y 1 6)2 5 49

26. 4x2 2 9y2 2 40x 1 64 5 0

A 5 4, B 5 0, C 5 29

B2 2 4AC 5 0 2 4(4)(29) 5 144


4x2 2 9y2 2 40x 1 64 5 0

4(x2 2 10x) 2 9y2 5 264

4(x2 2 10x 1 25) 2 9y2 5 264 1 100

4(x 2 5)2 2 9y2 5 36

(x 2 5)2

} 9 2

y2

} 4 5 1

27. y2 2 16y 2 12x 1 40 5 0

A 5 0, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(0)(1) 5 0


y2 2 16y 2 12x 1 40 5 0

y2 2 16y 5 12x 2 40

y2 2 16y 1 64 5 12x 2 40 1 64

( y 2 8)2 5 12x 1 24

( y 2 8)2 5 12(x 1 2)

28. 25x2 1 4y2 1 50x 2 24y 2 39 5 0

A 5 25, B 5 0, C 5 4

B2 2 4AC 5 0 2 4(25)(4) 5 2400


25x2 1 4y2 1 50x 2 24y 2 39 5 0

25(x2 1 2x) 5 4( y2 2 6y) 5 39

25(x2 1 2x 1 1) 1 4( y2 2 6y 1 9) 5 39 1 25 1 36

25(x 1 1)2 1 4( y 2 3)2 5 100

(x 1 1)2

} 4 1

( y 2 3)2

} 25 5 1

29. y2 2 16x2 1 14y 1 64x 2 31 5 0

A 5 216, B 5 0, C 5 1

B2 2 4AC 5 0 2 4(216)(1) 5 64


y2 2 16x2 1 14y 1 64x 2 31 5 0

( y2 1 14y) 2 16(x2 2 4) 5 31

( y2 1 14y 1 49) 2 16(x2 2 4x 1 4) 5 31 1 49 2 64

( y 1 7)2 2 16(x 2 2)2 5 16

( y 1 7)2

} 16

2 (x 2 2)2 5 1

30. 4x2 1 y2 5 16

x 1 y 5 2 → y 5 2 2 x

4x2 1 (2 2 x)2 5 16

4x2 1 4 2 4x 1 x2 5 16

5x2 2 4x 2 12 5 0

(5x 1 6)(x 2 2) 5 0

5x 5 26 or x 2 2 5 0

x 5 2 6 } 5 or x 5 2

When x 5 2 6 } 5 : y 5 2 2 1 2

6 } 5 2 5

16 } 5

When x 5 2: y 5 2 2 (2) 5 0

The solutions are 1 2 6 } 5 ,

16 } 5 2 and (2, 0).

31. x2 1 4y2 2 8y 5 4

y2 2 2y 2 8x 2 16 5 0 3 (24)

x2 1 4y2 2 8y 2 4 5 0

24y2 1 32x 1 8y 1 64 5 0

x2 1 32x 1 60 5 0

(x 1 2)(x 1 30) 5 0

x 5 22 or x 5 230

When x 5 22: When x 5 230:

(22)2 1 4y2 2 8y 5 4 (230)2 1 4y2 2 8y 5 4

4y2 2 8y 5 0 4y2 2 8y 1 896 5 0

4y(y 2 2) 5 0 4 1 y2 2 2y 1 224 2 5 0

y 5 0 or y 5 2 No real solution


32. 2x2 1 y2 1 2x 2 5 5 0

x2 1 y2 2 2x 2 3 5 0

2y2 2 8 5 0

2( y2 2 4) 5 0

y2 5 4

y 5 62

When y 5 62: (62)2 2 x2 1 2x 2 5 5 0

2x2 1 2x 2 1 5 0

2(x2 2 2x 1 1) 5 0

2(x 2 1)2 5 0

x 5 1



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33. Upright glass:

radius 5 1.5, r2 5 2.25

An equation of the water’s surface with the glass upright is x2 1 y2 5 2.25.

Tilted glass:

The water’s edge forms an ellipse.

a 5 1 } 2 (6) 5 3, a2 5 9

b 5 1 } 2 (3) 5

3 } 2 , b2 5

9 } 4

An equation is x2

} 9 1

4y2

} 9 5 1.

34. Parabola with point (2104, 45)

The form of the equation is x2 5 4py.

When x 5 2104 and y 5 45:

(2104)2 5 4p(45)

10,816 5 4p(45)

4p 5 10,816

} 45

An equation is x2 5 10,816

} 45 y.

4p 5 10,816

} 45

p 5 2704

} 45 ø 60.1

The distance from the vertex to the focus is about 60.1.

Standardized Test Preparation (p. 675)

1. Partial credit would be given, because the solution was without error but incomplete. Part of the question was not answered.

2. Full credit would be given, because the solution is complete and correct.

Standardized Test Practice (pp. 676–677)

1. A(6, 3), B(23, 5), C(25, 24) and D(4, 25)

Distances between the points:

AB 5 Ï}}

(6 2 (23))2 1 (3 2 5)2 5 Ï}

85

BC 5 Ï}}}

(23 2 (25))2 1 (5 2 (24))2 5 Ï}

85

CD 5 Ï}}}

(25 2 4)2 1 (24 2 (25))2 5 Ï}

81

DA 5 Ï}}

(4 2 6)2 1 (25 2 3)2 5 Ï}

68

The vertex D(4, 25) can be moved so that

CD 5 DA 5 Ï}

85 .

This can be done by drawing two circles around the points A(6, 3) and C(25, 24), each with a radius of Ï

}

85 , and fi nding their intersection, where x > 0 and y < 0.

The equations of the circles are:

(x 2 6)2 1 ( y 2 3)2 5 85

(x 1 5)2 1 ( y 1 4)2 5 85

x2 1 y2 2 12x 2 6y 2 40 5 0

2x2 2 y2 2 10x 2 8y 1 44 5 0

222x 2 14y 1 4 5 0

x 5 2 7 } 11 y 1

2 } 11

Substitute for x.

1 2 7 } 11 y 1

2 } 11 2 6 2 21 ( y 2 3)2 5 85

1 2 7 } 11 y 2

64 } 11 2 2 1 ( y 2 3)2 5 85

49

} 121

y2 1 896

} 121 y 1 4096

} 121 1 y2 2 6y 1 9 2 85 5 0

49y2 1 896y 1 4096 1 121y2 2 726y 2 9196 5 0

170y2 1 170y 2 5100 5 0

170( y2 1 y 2 30) 5 0

170( y 1 6)( y 2 5) 5 0

y 5 26 or y 5 5

Because y < 0 for vertex D, use y 5 26:

x 5 2 7 } 11 (26) 1

2 } 11 5

42 } 11 1

2 } 11 5

44 } 11 5 4

The vertex D should be (4, 26) to make ABCD a rhombus.

2. The form of the parabola’s equation is x2 5 4py.

When x 5 24 and y 5 212.5:

(24)2 5 4p(12.5)

p 5 (24)2

} (4)(212.5)

5 211.52

The focus is 11.52 inches below the vertex of the parabola. The focus is above the bottom edge of the refl ector, because 11.52 < 12.5.

3.

x

y

20

240

B (40, 80)

(235, 35)

O (0, 0)

(55, 25)A

Slope of } AO :

m 5 35 2 0

} 235 2 0 5 21

2 1 } m 5 1

Midpoint of } AO :

M1235 1 0}

2 , 35 1 0}

2 2 5 1235}2 ,

35}2 2

Perpendicular bisector of }AO :

y 2 35}2 5 1 1x 2 1235

}2 2 2

y 5 x 1 35}2 1

35}2

y 5 x 1 35


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599Algebra 2


Slope of } OB :

m 5 0 2 80

} 0 2 40 5 2

2 1 } m 5 2

1 } 2

Midpoint of } OB : M 1 0 1 40 } 2 ,

0 1 80 }

2 2 5 (20, 40)

Perpendicular bisector of } OB :

y 2 40 5 2 1 } 2 (x 2 20)

y 5 2 1 } 2 x 1 10 1 40

y 5 2 1 } 2 x 1 50


x 1 35 5 2 1 } 2 x 1 50

3 }

2 x 5 15

x 5 10

When x 5 10: y 5 (10) 1 35 5 45

The center of the crater is at (10, 45).

Radius 5 Ï}}

(10 2 0)2 1 (45 2 0)2

5 Ï}

100 1 2025 5 Ï}

2125

An equation for the edge of the crater is

(x 2 10)2 1 (y 2 45)2 5 2125.

Test point (55, 25):

(55 2 10)2 1 (25 2 45)2 ≤? 2125

2025 1 400 ≤? 2125

2425 ÷ 2125

The point (55, 25) is outside of the crater, because it is 2425 kilometers from the crater’s center and the crater’s radius is 2125 kilometers.

4. The sum of the distances from any red point to each of the foci is always 10.

The vertices are 10 units apart, so 2a 5 10.

a 5 1 } 2 (10) 5 5

a2 5 25

The foci are 7 units apart, so 2c 5 7.

c 5 7 } 2

c2 5 49

} 4

c2 5 a2 2 b2

49

} 4 5 25 2 b2

b2 5 25 2 49

} 4 5 51

} 4 5 12.75

An equation is x2

} 25

1 y2

} 12.75 5 1.

5. The circle’s center is at (0, 212) with a radius of 10. Its equation is x2 1 ( y 1 12)2 5 102.

The penalty area’s edge intersects the circle wheny 5 218:

x2 1 (218 1 12)2 5 102

x2 1 36 5 100

x2 5 64

x 5 68

Distance between the points (28, 218) and (28, 218):

d 5 Ï}}}

(28 2 8)2 1 (218 2 (218))2 5 Ï}

162 5 16

The distance along the edge of the penalty area between the endpoints of the arc is 16 yards.

6.

x

y

x

y

x

y

x

y

No solutions 4 solutions

7. The hyperbola’s vertices are (22.6, 0) and (2.6, 0).

So a 5 2.6, a2 5 6.76

The point (6.3, 18.2) lies on the hyperbola, so

(6.3)2

} (2.6)2 2

(18.2)2

} b2 5 1

2 (18.2)2

} b2 ø 24.87

b2 ø 68.02

An equation for the hyperbola is x2 }

6.76 2

y2

} 68.02 5 1.

If the origin is moved to the bottom left of the sculpture, the center will move up and to the right to (6.3, 18.2).

With h 5 6.3 and k 5 18.2, the new equation will be

(x 2 6.3)2

} 6.76

2 ( y 2 18.2)2

} 68.02 5 1.

8. 4x2 1 6y2 5 48 → x2

} 12

1 y2

} 8 5 1 and

(x 2 5)2

} 12

1 (y 1 7)2

} 8 5 1

The two ellipses have the same area because they are the same size and shape. The only difference is that they have different centers.

9. x2 2 y2 1 4x 2 18y 2 15 5 0

A 5 1, B 5 0, C 5 21

B2 2 4AC 5 0 2 4(1)(21) 5 4

B2 2 4AC > 0, so the equation represents a hyperbola.

10. A;

x2 1 y2 5 36

2y2 1 x 5 6

x2 1 x 5 42

x2 1 x 2 42 5 0

(x 1 7)(x 2 6) 5 0

x 5 27 or x 5 6


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When x 5 27: y2 5 2(27)2 1 36

y2 5 249 1 36 5 213

No real solution

When x 5 6: y2 5 2(6)2 1 36

y2 5 236 1 36 5 0

y 5 0


11. D; 12(x 2 6) 5 2( y 1 4)2

The graph is a parabola that opens left, so the graph does not represent a function.

12. (x 2 2)2

} 81

2 (y 2 5)2

} 4 5 1

Line of symmetry: x 5 s

Because h 5 2, and k 5 5, the center of the hyperbola is (2, 5). Therefore, its line of symmetry is x 5 2.

13. (210, 2), (7, 7)

d 5 Ï}}

(210 2 7)2 1 (2 2 7)2

5 Ï}}

(217)2 1 (25)2

ø 17.7

14. 9x2 2 4y2 5 324

x2

} 36

2 y2

} 81 5 1

a2 5 36, a 5 6

The length of the transverse axis is 2a 5 12.

15. 8x2 1 8y2 5 720

x2 1 y2 5 90

r2 5 90

r ø 9.5

16. Circumference

π F 3a 1 3b 2 Ï}}

(a 1 3b)(b 1 3a) G Ellipse with center (0, 0); vertex (0, 25); focus (0, 3)

a 5 5, c 5 3

c2 5 a2 2 b2

(3)2 5 (5)2 2 b2

b2 5 25 2 9 5 16

b 5 4

Circumference

π F 3(5) 1 3(4) 2 Ï}}

(5 1 3(4))(4 1 3(5)) G 5 π(15 1 12 2 Ï

}

(17)(19) ) ø π(27 2 17.97)

ø 28

The circumference of the ellipse is approximately 28 units.

17. a. An equation of the original orbit is x2 1 y2 5 42002.

An equation of the geostationary orbit isx2 1 y2 5 22,2402.

b. For the transfer orbit, a 2 c is equal to the radius of the original orbit. a 2 c 5 4200

For the transfer orbit, a 1 c is equal to the radius of the geostationary orbit. a 1 c 5 22,400

By solving the system of equations, the values of a and c can be determined.

a 2 c 5 4200

a 1 c 5 22,240

2a 5 26,440

a 5 13,220

When a 5 13,220: a 2 c 5 4200

13,220 2 c 5 4200

c 5 13,220 2 4200

c 5 9020

c. The center of the transfer orbit is c 5 9020 miles to the right of Earth’s center. Center: (9020, 0)

h 5 9020, k 5 0

c2 5 a2 2 b2

(9020)2 5 (13,220)2 2 b2

b2 5 (13,220)2 2 (9020)2

b2 5 93,408,000

b ø 9665

An equation of the ellipse is

(x 2 9020)2

} 13,2202 1

y2

} 96652 5 1.

18. a. Find the midpoint of each side. Then fi nd the midpoint of that point and the closer vertex.

b. For the blue square:

x

y

8

8

A

B

C

D

(0, 32)

(32, 48)

(48, 16)

(16, 0)

E

F

G

H

M1

M2

M3

M4

M1 1 0 1 32 }

2 ,

32 1 48 }

2 2 5 (16, 40)

E 1 16 1 32 }

2 ,

40 1 48 }

2 2 5 (24, 44)

M2 1 32 1 48 }

2 ,

48 1 16 }

2 2 5 (40, 32)

F 1 40 1 48 }

2 ,

32 1 16 }

2 2 5 (44, 24)

M3 1 48 1 16 }

2 ,

16 1 0 }

2 2 5 (32, 8)

G 1 32 1 16 }

2 ,

8 1 0 }

2 2 5 (24, 4)

M4 1 16 1 0 }

2 ,

0 1 32 }

2 2 5 (8, 16)

H 1 8 1 0 }

2 ,

16 1 32 }

2 2 5 (4, 24)


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601Algebra 2


The vertices of the blue square are (24, 44), (44, 24), (24, 4), and (4, 24).

For the green square:

x

y

8

8

A

B

C

D

(0, 32)

(32, 48)

(48, 16)

(16, 0)

E

F

G

H

M1 M2

M3M4

WX

YZ

M1 1 4 1 24 }

2 ,

24 1 44 }

2 2 5 (14, 34)

W 1 14 1 24 }

2 ,

34 1 44 }

2 2 5 (19, 39)

M2 1 24 1 44 }

2 ,

44 1 24 }

2 2 5 (34, 34)

X 1 34 1 44 }

2 ,

34 1 24 }

2 2 5 (39, 29)

M3 1 44 1 24 }

2 ,

24 1 4 }

2 2 5 (34, 14)

Y 1 34 1 24 }

2 ,

14 1 4 }

2 2 5 (29, 9)

M4 1 24 1 4 }

2 ,

4 1 24 }

2 2 5 (14, 14)

Z 1 14 1 4 }

2 ,

14 1 24 }

2 2 5 (9, 19)

The vertices of the green square are (19, 39), (39, 29), (29, 9), and (9, 19).

c. Red square:

Side length 5 Ï}}

(32 2 0)2 1 (48 2 32)2 5 Ï}

1280

Area 5 ( Ï}

1280 )2 5 1280

Blue square:

Side length 5 Ï}}

(24 2 4)2 1 (44 2 24)2 5 Ï}

800

Area 5 ( Ï}

800 )2 5 800

Green square:

Side length 5 Ï}}

(19 2 9)2 1 (39 2 19)2 5 Ï}

500

Area 5 ( Ï}

500 )2 5 500

The area of each square is 5 }

8 the area of the previous

square in the pattern.

Cumulative Review Chs. 1–9 (pp. 678–679)

1. 5x 1 24 5 11 2 2x

7x 5 213

x 5 2 13

} 7

Check: 5 1 2 13

} 7 2 1 24 0 11 2 2 1 2 13

} 7 2

2 65

} 7 1 24 0 11 1 26

} 7

103

} 7 5 103

} 7 ✓

2. 4x 2 7 5 13

4x 5 20 or 4x 5 26

x 5 5 or x 5 2 3 } 2

Check:

4(5) 2 7 0 13 4 1 2 3 } 2 2 2 7 0 13

13 0 13 2 26

} 2 0 13

13 5 13 ✓ 13 5 13 ✓

3. x2 2 12x 1 35 5 0

(x 2 7)(x 2 5) 5 0

x 5 7 or x 5 5

Check:

72 2 12(7) 1 35 0 0 52 2 12(5) 0 0

49 2 84 1 35 0 0 25 2 60 1 35 0 0

0 5 0 ✓ 0 5 0 ✓

4. 2x2 2 5x 1 5 5 0

x 5 2 (25) 6 Ï

}}

(25)2 2 4(2)(5) }}}

2(2)

x 5 5 6 Ï

}

215 } 4 5

5 6 i Ï}

15 } 4

Check:

2 1 5 1 i Ï}

15 }

4 2 2 2 5 1 5 1 i Ï

}

15 }

4 2 1 5 0 0

2 1 5 1 5i Ï}

15 }

8 2 2

25 } 4 2

5i Ï}

15 } 4 1 5 0 0

5 }

4 1

5i Ï}

15 } 4 2

25 } 4 2 5i Ï

}

15 1 5 0 0

0 5 0 ✓

2 1 5 2 i Ï}

15 }

4 2 2 2 5 1 5 2 i Ï

}

15 }

4 2 1 5 0 0

2 1 5 2 5i Ï}

15 }

8 2 2

25 } 4 1

5i Ï}

15 } 4 1 5 0 0

5 }

4 2

5i Ï}

15 } 4 2

25 } 4 1

5i Ï}

15 } 4 1 5 0 0

0 5 0 ✓

5. x3 1 3x2 2 18x 5 40

x3 1 3x2 2 18x 2 40 5 0

(x 1 2)(x2 1 x 2 20) 5 0

(x 1 2)(x 1 5)(x 2 4) 5 0

x 5 22 or x 5 25 or x 5 4

Check:

(22)3 1 3(22)2 2 18(22) 0 40

28 1 12 1 36 0 40

40 5 40 ✓

(25)3 1 3(25)2 2 18(25) 0 40

2125 1 75 1 90 0 40

40 5 40 ✓


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43 1 3(4)2 2 18(4) 0 40

64 1 48 2 72 0 40

40 5 40 ✓

6. Ï}

x 2 2 5 x 2 4

x 2 2 5 (x 2 4)2

x 2 2 5 x2 2 8x 1 16

x2 2 8x 1 16 2 x 1 2 5 0

x2 2 9x 1 18 5 0

(x 2 3)(x 2 6) 5 0

x 5 3 or x 5 6

Check:

Ï}

3 2 2 0 3 2 4 Ï}

6 2 2 0 6 2 4

Ï}

1 0 21 Ï}

4 0 2

1 Þ 21 2 5 2 ✓

Reject x 5 3. The solution is 6.

7. 4x 2 5 5 3

4x 5 8

(22)x 5 23

22x 5 23

2x 5 3

x 5 3 } 2

Check:

43/2 2 5 0 3

8 2 5 0 3

3 5 3 ✓

8. x 1 3

} 3x 1 1

5 x } x 1 2

(x 1 3)(x 1 2) 5 x(3x 1 1)

x2 1 5x 1 6 5 3x2 1 x

2x2 2 4x 2 6 5 0

2(x2 2 2x 2 3) 5 0

2(x 1 1)(x 2 3) 5 0

x 5 21 or x 5 3

Check:

21 1 3

} 3(21) 1 1

0 21 } 21 1 2

3 1 3

} 3(3) 1 1

0 3 }

3 1 2

2 }

22 0 21 }

1

6 }

10 0

3 }

5

21 5 21 ✓ 3 } 5 5

3 } 5 ✓

9. x 2 4

} x 2 3

1 2 5 2x 2 3

} x 2 3

(x 2 3) 1 x 2 4 }

x 2 3 2 1 (x 2 3)2 5 (x 2 3) 1 2x 2 3

} x 2 3

2 x 2 4 1 2x 2 6 5 2x 2 3

x 2 7 5 0

x 5 7

Check:

7 2 4

} 7 2 3

1 2 0 2(7) 2 3

} 7 2 3

3 }

4 1 2 0 11 }

4

11

} 4 5

11 } 4 ✓

10. y 5 22x 1 7 11. y 5 (x 1 1)2(x 2 2)

x

y

22

2

1

x

y

22

12. y 5 Ï}

x 1 4 1 3 13. y 5 4ex

x

y

21

1

(24, 3)

5

x

y

21

14. y 5 ln(x 2 2) 15. y 5 3x 2 1

} x2 2 9

1

x

y

x 5 2

21

1

x

y

21

x 5 3

x 5 23

16. 2x2 2 20x 2 48 5 2(x2 2 10x 2 24) 5 2(x 1 2)(x 2 12)

17. 6x2 1 7x 2 20 5 (3x 2 4)(2x 1 5)

18. x3 1 8x2 2 4x 2 32 5 (x 1 8)(x2 2 4) 5 (x 1 8)(x 1 2)(x 2 2)

19. f (x) 5 6x 2 1

y 5 6x 2 1

6x 5 y 1 1

x 5 y 1 1

} 6

Write the inverse as y 5 x 1 1

} 6 .

The inverse function is f (x) 5 x 1 1

} 6 .


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603Algebra 2


20. f (x) 5 x3 2 5

y 5 x3 2 5

x3 5 y 1 5

x 5 3 Ï}

x 1 5

Write the inverse function as y 5 3 Ï}

x 1 5 .

The inverse function is f (x) 5 3 Ï}

x 1 5 .

21. f (x) 5 x5

y 5 x5

x 5 y1/5

Write the inverse function as y 5 x1/5.

The inverse function is f (x) 5 x1/5.

22. f (x) 5 5(1.4x) The function is an example of exponential growth

because 1.4 > 1.

23. f (x) 5 3(0.6)x

The function is an example of exponential decay because 0.6 < 1.

24. f (x) 5 8e22x

The function is an example of exponential decay because e22 < 1.

25. 3 ln x 2 ln 5 5 ln x3 1 ln 521

5 ln x3

} 5

26. log3 4 1 2 log3 7 5 log3 4 1 log3 72

5 log3(4)(49)

5 log3 196

27. 5 log x 1 log y 2 3 log z 5 log x5 1 log y 1 log z23

5 log x5y

} z3

28. x 5 18, y 5 6 29. x 5 5, y 5 215

y 5 a } x y 5

a } x

6 5 a } 18 (215) 5

a } 5

a 5 108 275 5 a

y 5 108

} x y 5 2 75

} x

30. x 5 6, y 5 9

y 5 a } x

9 5 a } 6

54 5 a

y 5 54

} x

31. x 2 5

} x 1 7 : 3x 1 21

} x2 2 25

5 (x 2 5)(3)(x 1 7)

}} (x 1 7)(x 1 5)(x 2 5)

5 3 } x 1 5

32. 2x 1 8

} x 2 3

4 x 1 4

} x2 2 x 2 6

5 2(x 1 4)(x 1 2)(x 2 3)

}} (x 2 3)(x 1 4)

5 2(x 1 2) 5 2x 1 4

33. x 2 3

} x 1 5 1 7 } x 2 2

5 (x 2 3)(x 2 2)

}} (x 1 5)(x 2 2)

1 7(x 1 5)

}} (x 1 5)(x 2 2)

5 x2 2 5x 1 6

}} (x 1 5)(x 2 2)

1 7x 1 35

}} (x 1 5)(x 2 2)

5 x2 1 2x 1 41

} x2 1 3x 2 10

34. (28, 5), (24, 21)

d 5 Ï}}}

(28 2 (24))2 1 (5 2 (21))2

5 Ï}

(24)2 1 62

5 Ï}

16 1 36

5 Ï}

52

5 2 Ï}

13

M 1 28 1 (24) } 2 ,

5 1 (21) }

2 2 5 1 212

} 2 , 4 }

2 2 5 (26, 2)

35. (3, 5), (8, 7)

d 5 Ï}}

(3 2 8)2 1 (5 2 7)2

5 Ï}}

(25)2 1 (22) 2

5 Ï}

25 1 4

5 Ï}

29

M 1 3 1 8 } 2 ,

5 1 7 }

2 2 5 1 11

} 2 , 6 2 36. (22, 7), (1, 14)

d 5 Ï}}

(22 2 1)2 1 (7 2 14)2

5 Ï}}

(23)2 1 (27)2

5 Ï}

9 1 49

5 Ï}

58

M 1 22 1 1 } 2 ,

7 1 14 }

2 2 5 1 2

1 } 2 ,

21 }

2 2

37. x2 1 y2 1 12x 2 4y 1 15 5 0

A 5 1, B 5 0, C 5 1

B2 2 4AC 5 24

The conic is a circle because B2 2 4AC < 0, B 5 0, andA 5 C.

x2 1 y2 1 12x 2 4y 1 15 5 0

(x2 1 12x) 1 ( y2 2 4y) 5 215

(x2 1 12x 1 36) 1 ( y2 2 4y 1 4) 5 215 1 36 1 4

(x 1 6)2 1 (y 2 2)2 5 25

h 5 26, k 5 2

r2 5 25, r 5 5


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Center: (26, 2)

6

x

y

24

(26, 2)

38. 4x2 2 16y2 2 56x 1 160y 2 268 5 0

A 5 4, B 5 0, C 5 216

B2 2 4AC 5 24(4)(216) 5 256


4x2 2 16y2 2 56x 1 160y 2 268 5 0

4 (x2 2 14) 2 16 ( y2 2 10y) 5 268

4 (x2 2 14x 1 49) 2 16 ( y2 2 10y 1 25) 5 268 1 196 2 400

4 (x 2 7)2 2 16(y 2 5)2 5 64

(x 2 7)2

} 16

2 (y 2 5)2

} 4 5 1

h 5 7, k 5 5

Center: (7, 5)

a2 5 16 a 5 4

b2 5 4 b 5 2

c2 5 a2 1 b2 5 20

The vertices are 4 units left and right of the center, at(3, 5) and (11, 5).

4

x

y

22

(7, 7)

(7, 5)(7, 3)

(3, 5)(11,5)

39. y2 1 6x 1 4y 1 16 5 0

A 5 0, B 5 0, C 5 1


y2 1 6x 1 4y 1 16 5 0

y2 1 4y 5 26x 2 16

y2 1 4y 1 4 5 26x 2 16 1 4

(y 1 2)2 5 26(x 1 2)

h 5 22, k 5 22

Vertex: (22, 22)

4p 5 26

p 5 2 3 } 2

The focus is 2 3 } 2 units left of the vertex, at 1 2

7 } 2 , 22 2 .

2

x

y

2

(22, 22)

( )72 , 222

40. 2x2 1 3y2 1 4x 1 12y 2 14 5 0

A 5 2, B 5 0, C 5 3

B2 2 4AC 5 24(2)(3) 5 224


2x2 1 3y2 1 4x 1 12y 2 14 5 0

2 1 x2 1 2x 2 1 3 1 y2 1 4y 2 5 14

2 1 x2 1 2x 1 1 2 1 3 1 y2 1 4y 1 4 2 5 14 1 2 1 12

2(x 1 1)2 1 3(y 1 2)2 5 28

(x 1 1)2

} 14

1 (y 1 2)2

} 28

} 3 5 1

h 5 21, k 5 22

Center: (21, 22)

a2 5 14 a 5 Ï}

14

b2 5 28

} 3 b 5 Ï}

28

} 3

c2 5 a2 2 b2

c2 5 14 2 28

} 3 5 14

} 3

c 5 Ï}

14

} 3

The vertices are Ï}

14 units left and right of the center, at

(21 2 Ï}

14 , 22) and (21 1 Ï}

14 , 22).

The co-vertices are Ï}

28

} 3 units above and below the

center, at 1 21, 22 1 Ï}

28

} 3 2 and 1 21, 22 2 Ï

}

28

} 3 2 .

4

x

y

22

14, 22212( ) 14, 22211( )

( )28321, 22 1

( )28321, 22 2

(21, 22)


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605Algebra 2


41.

x

y

x 5 y 1 40

2x 1 2y 5 380

2( y 1 40) 1 2y 5 380

2y 1 80 1 2y 5 380

4y 5 300

y 5 75

x 5 (75) 1 40 5 115

The width of the garden is 75 feet and the length is 115 feet.

42. g 5 games

r 5 rides

1.5g 1 2.5r ≤ 30

1.5g ≤ 30 2 2.5r

g ≤ 2 } 3 1 2 5 } 2 r 2 1 30 1 2 } 3 2

g ≤ 2 5 } 3 r 1 20

An inequality representing the number of games and rides that can be purchased for $30 or less is

g ≤ 2 5 } 3 r 1 20.

Number of rides

Nu

mb

er o

f g

ames

0 2 6 10 144 8 12 r

g

0

3

9

15

21

6

12

18(0, 20)

(12, 0)

The ordered pairs (rides, games) that use exactly $30 are (0, 20), (3, 15), (6, 10), (9, 5) and (12, 0).

43. Adultprice p

Number of adults 1

Studentprice p

5 p x 1 2 p Number

of students 5 Total

income (650 2 x) 5 2500

5x 1 1300 2 2x 5 2500

3x 5 1200

x 5 400

There were 400 adults and 650 2 400 5 250 students at the basketball game.

44. T 5 F 0.95 0.02 0.05 0.98G M0 5 F 700

700G M1 5 TM0 5 F 0.95 0.02

0.05 0.98G F 700 700G

5 F 665 1 14 35 1 686G 5 F 679

721G M2 5 TM1 5 F 0.95 0.02

0.05 0.98G F 679 721G

5 F 645 1 14 34 1 707G 5 F 659

741G M3 5 TM2 5 F 0.95 0.02

0.05 0.98G 5 F 659 741G

5 F 626 1 15 33 1 726G 5 F 641

759G The matrices M1, M2, and M3 represent the distribution

of customers in company A and company B for the fi rst three months. Each month, the total number of customers for company A is declining, while the total number of customers for company B is increasing.

45. Area 5 6 1 } 2

2 2 1 12 19 1

18 7 1 5 6

1 } 2

2 2 1 12 19 1

18 7 1 2 2

12 19

18 7

5 6 1 } 2 [(38 1 36 1 84) 2 (342 1 14 1 24)]

5 6 1 } 2 (2222)

5 111

111 units2 p 1 0.25mi }

1 unit 2 2 ø 6.9 mi2

The area of the surface of the lake is about 6.9 square miles.

46. h 5 216t2 1 v0t 1 h0

The ball’s initial velocity is v0 5 16 feet per second.

The initial height of the ball is h0 5 9 feet.

Let h 5 0 and solve for t to fi nd the time when the ball hits the ground.

0 5 216t2 1 16t 1 9

t 5 216 6 Ï

}}

(16)2 2 4(216)(9) }}}

2(216)

t ø 216 6 28.8 }

232

t 5 1.4 or 20.4

The ball hits the ground in about 1.4 seconds.


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47. The function that gives the sale price of the television after $50 is subtracted and a 15% discount is applied is

f (t) 5 0.85(t 2 50).

If t 5 $480:

f (t) 5 0.85(480 2 50)

5 0.85(430)

5 365.5

The sale price of the television will be $365.50.

48. p 5 4500

r 5 0.0275

n 5 12

A 5 P 1 1 1 r } n 2

nt

A 5 4500 1 1 1 0.0275

} 12 2 12t

A 5 4500(1.0023)12t

When t 5 5:

A 5 4500(1.0023)60 ø 5165.07

The balance after 5 years will be $5165.07.

49. t 0 10 20 30 40 50

ln s 5.75 6.13 6.42 6.62 6.83 6.97

t 60 70 80 90 100

ln s 7.14 7.35 7.65 7.97 8.27

The data pairs (t, ln s) lie close to a line.

m 5 8.27 2 5.75

} 100 2 0 ø 0.025

ln s 2 5.75 5 0.025(t 2 0)

ln s 5 0.025t 1 5.75

s 5 e0.025t 1 5.75

s 5 e5.75 1 e0.025 2 t

s 5 314(1.025)t

An exponential model for the data pairs (t, s) iss 5 314(1.025)t.

Years since 1904

0 20 40 60 80 100 t

n s

0

5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

9.0

Cu

mu

lati

ve n

um

ber

of

stam

p d

esig

ns

(100, 8.27)

(0, 5.75)


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Mc D Chapter 9 Solutions1

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Transcript of Mc D Chapter 9 Solutions1