eSolutions Manual - Powered by Cognero Page 1 · foci: (h, k c) = (0, 2.83) and ( 0, í2.83 ) Graph...

Graph the hyperbola given by each equation.

1. = 1

SOLUTION:The equation is in standard form, where h = 0, k =

0, a = or4,b = or3,andc = or5. In the standard form of the equation, the y-termisbeing subtracted. Therefore, the orientation of the hyperbola is horizontal. center: (h, k) = (0, 0)

vertices: (h a, k) = (4, 0) and (4, 0) foci: (h c, k) = (5, 0) and (5, 0)

Graph the center, vertices, foci, and asymptotes. Then make a table of values to sketch the hyperbola.

2. = 1

SOLUTION:

The equation is in standard form, where h = 0, k =

b = orabout4.12,andc = orabout standard form of the equation, the x-term is being sTherefore, the orientation of the hyperbola is vertic center: (h, k) = (0, 0)

vertices: (h, k a) = (0, 2) and (0 2) foci: (h, k c) = (0, 4.58) and (0, 4.58)

Graph the center, vertices, foci, and asymptotes. Thvalues to sketch the hyperbola.

3. = 1


0, a = or7,b = orabout5.48,andc =

orabout8.89.Inthestandardformofthe equation, the y-termisbeingsubtracted.Therefore, the orientation of the hyperbola is horizontal. center: (h, k) = (0, 0)

vertices: (h a, k) = (7, 0) and (7, 0) foci: (h c, k) = (8.89, 0) and (0, 8.89)

Graph the center, vertices, foci, and asymptotes. Then use a table of values to sketch the hyperbola.

4. = 1

SOLUTION:


about 5.83, b = orabout3.74,andc = 6.93. In the standard form of the equation, the x-tersubtracted. Therefore, the orientation of the hyperb center: (h, k) = (0, 0)

vertices: (h, k a) = (0, 5.83) and (0, 5.83) foci: (h, k c) = (0, 6.93) and (0, 6.93)


5. = 1



orabout5.48.Inthestandardformoftheequation, the y-term is being subtracted. Therefore,the orientation of the hyperbola is horizontal. center: (h, k) = (0, 0)

vertices: (h a, k) = (3, 0) and (3, 0 ) foci: (h c, k) = (5.48, 0) and (5.48, 0)


6. =1


0, a = or6,b = or2,andc =

orabout6.32.Inthestandardformoftheequation, the yterm is being subtracted. Therefore,the orientation of the hyperbola is horizontal. center: (h, k) = (0, 0)



7. = 1


or9,b = orabout2.83,andc = 9.43. In the standard form of the equation, the xtesubtracted. Therefore, the orientation of the hyperbvertical. center: (h, k) = (0, 0)

vertices: (h, k a) = (0, 9) and (0, 9) foci: (h, k c) = (0, 9.43) and (0, 9.43)

Graph the center, vertices, foci, and asymptotes. Thtable of values to sketch the hyperbola.

8. = 1

SOLUTION:


b = orabout3.74,andc = oraboustandard form of the equation, the xterm is being sTherefore, the orientation of the hyperbola is vertic center: (h, k) = (0, 0)



9.3x2 2y2 = 12

SOLUTION:First, divide each side by 12 to write the equation instandard form.

The equation is now in standard form, where h = 0,

k = 0, a = or2,b = orabout2.45,andc =




10.3y2 5x2 = 15

SOLUTION:First, divide each side by 15 to write the equation inform.


orabout2.24,b = orabout1.73,andc = about 2.83. In the standard form of the equation, thbeing subtracted. Therefore, the orientation of the hvertical. center: (h, k) = (0, 0)



11.LIGHTINGThelightprojectedonawallbyatable lamp can be represented by a hyperbola. The light from a certain table lamp can be modeled by

=1. Graph the hyperbola. Refer to the

photo on Page 449.


0, a = or15,b = or9,andc =

orabout 17.49. In the standard form ofthe equation, the xterm is being subtracted. Therefore, the orientation of the hyperbola is vertical. center: (h, k) = (0, 0)




12. = 1



orabout 7.55. In the standard form of theequation, the yterm is being subtracted. Therefore,the orientation of the hyperbola is horizontal. center: (h, k) = (5, 4) vertices: (h a, k) = (2, 4) and (8, 4) foci: (h c, k) = (2.55, 4) and (12.55, 4) asymptotes:


13. = 1



orabout 6.08. In the standard form of theequation, the xterm is being subtracted. Therefore,the orientation of the hyperbola is vertical. center: (h, k) = (0, 7)

vertices: (h, k a) = (0, 9) and (0, 5) foci: (h, k c) = (0, 13.08) and (0, 0.92) asymptote:


14. = 1



orabout 9.22. In the standard form of the equation, the yterm is being subtracted. Therefore, the orientation of the hyperbola is horizontal. center: (h, k) = (2, 6) vertices: (h a, k) = (7, 6) and (3, 6) foci: (h c, k) = (11.22, 6) and (7.22, 6) asymptotes:


15. = 1





16. = 1



orabout 7.92. In the standard form of the equation, the xterm is being subtracted. Therefore, the orientation of the hyperbola is vertical. center: (h, k) = (4, 3)

vertices: (h, k a) = (4, 7) and (4, 1) foci: (h, k c) = (4, 10.62) and (4, 4.62) asymptotes:


17. = 1





18.x2 4y2 6x 8y = 27

SOLUTION:First, write the equation in standard form.


k = 1, a = orabout6.32,b = orabout

3.16, and c = orabout 7.07. In the standard form of the equation, the yterm is being subtracted. Therefore, the orientation of the hyperbola is horizontal. center: (h, k) = (3, 1) vertices: (h a, k) = (9.32, 1) and (3.32, 1) foci: (h c, k) = (10.07, 1) and (4.07, 1) asymptotes:


19.x2 + 3y2 4x + 6y = 28


The equation is now in standard form, where h =

2, k = 1, a = or3,b = orabout5.2,

and c = or6. In the standard form of the equation, the xterm is being subtracted. Therefore,the orientation of the hyperbola is vertical. center: (h, k) = (2, 1) vertices: (h, k a) = (2, 2) and (2, 4) foci: (h, k c) = (2, 5) and (2, 7) asymptotes:


20.13x2 2y2 + 208x + 16y = 748



8, k = 4, a = or2,b = orabout5.1,and

c = orabout 5.5. In the standard form of the equation, the yterm is being subtracted. Therefore, the orientation of the hyperbola is horizontal. center: (h, k) = (8, 4) vertices: (h a, k) = (6, 4) and (10, 4) foci: (h c, k) = (2.5, 4) and (13.5, 4) asymptotes:


21.5x2 + 2y2 70x 8y = 287



7, k = 2, a = or5,b = orabout3.2,

and c = orabout 5.92. In the standard form of the equation, the xterm is being subtracted. Therefore, the orientation of the hyperbola is vertical. center: (h, k) = (7, 2) vertices: (h, k a) = (7, 7) and (7, 3) foci: (h, k c) = (7, 7.92) and (7, 3.92) asymptotes:


22.EARTHQUAKESShortlyafteraseismographdetects an earthquake, a second seismograph positioned due north of the first detects the earthquake. The epicenter of the earthquake lies ona branch of the hyperbola represented by

=1, where the

seismographs are located at the foci. Graph the hyperbola.


30, a = or30,b = or40,andc =

or50. In the standard form of the equation, the xterm is being subtracted. Therefore,the orientation of the hyperbola is vertical. center: (h, k) = (60, 30)

vertices: (h, k a) = (60, 60) and (60, 0) foci: (h, k c) = (60, 80) and (60, 20) asymptotes:


Write an equation for the hyperbola with the given characteristics.

23.foci (1, 9), (1, 7); conjugate axis length of 14 units

SOLUTION:Because the xcoordinates of the foci are the same, the transverse axis is vertical, and the

standard form of the equation is

=1.

The center is the midpoint of the segment between

the foci, or (1, 1). So, h = 1 and k = 1. The length of the conjugate axis of a hyperbola is 2b.

So, 2b = 14, b = 7, and b2 = 49. You can find c by

determining the distance from a focus to the center.One focus is located at (1, 9), which is 8 units from (1, 1). So, c = 8. Now you can use the values of b and c to find a.

Using the values of h, k , a, and b, the equation for

the hyperbola is =1.

a2 = c2 b2

a2 = 82 72

a2 = 64 49

a2 = 15

a =

24.vertices (7, 5), (5, 5); foci (11, 5), (9, 5)

SOLUTION:Because the ycoordinates of the vertices are the same, the transverse axis is horizontal, and the


=1.

The center is the midpoint of the segment between the foci, or (1, 5). So, h = 1 and k = 5. You can findc by determining the distance from a focus to the center. One focus is located at (11, 5), which is 10 units from (1, 5). So, c = 10. You can find a by determining the distance from a vertex to the center. One vertex is located at (7, 5), which is 6

units from (1, 5). So, a = 6 and a2 = 36.

Now you can use the values of c and a to find b.



b2 = c

2 a

2

b2 = 10

2 6

2

b2 = 100 36

b2 = 64

b = 8


SOLUTION:Because the ycoordinates of the foci are the same, the transverse axis is horizontal, and the


=1.




determining the distance from a focus to the center.

One focus is located at (9, 1), which is 6 units from (3, 1). So, c = 6. Now you can use the values of b and c to find a.



a2 = c2 b2

a2 = 62 32

a2 = 36 9

a2 = 27

a =

26.vertices (1, 9), (1, 3); asymptotes y = x +

SOLUTION:Because the xcoordinates of the vertices are the same, the transverse axis is vertical, and the


=1.


the vertices, or (1, 6). So, h = 1 and k = 6. To find a, determine the distance from a vertex to the center. One vertex is located at (1, 9), which is 3

units from (1, 6). So, a = 3 and a2 = 9. Because

the slopes of the asymptotes are , using the

positive slope , b = 7 and b2 = 49.



27.vertices (3, 12), (3, 4); foci (3, 15), (3, 1)



=1.


the foci, or (3, 8). So, h = 3 and k = 8. You can find c by determining the distance from a focus

to the center. One focus is located at (3, 15) which is 7 units from (3, 8). So, c = 7. You can find a by determining the distance from a vertex to

the center. One vertex is located at (3, 12)

which is 4 units from (3, 8). So, a = 4 and a2 =

16. Now you can use the values of c and a find b.



b2 = c2 a2

b2 = 72 42

b2 = 49 16

b2 = 33

b =

28.foci (9, 7), (17, 7); asymptotes y = x +



=1.


the foci, or (4, 7). So, h = 4 and k = 7. The

slopes of the asymptotes are . So, the positive

slope isequalto , where b = 5, b2 = 25, a =

12, and a2 = 144.



29.center (7, 2); asymptotes y = x + ,

transverse axis length of 10 units

SOLUTION:

The center is (7, 2). Therefore, h = 7 and k = 2. The transverse axis has a length of 2a units. So, 10

= 2a, a = 5, and a2 = 25. The slopes of the

asymptotes are , and the standard form of the

equation is =1. So, the

positive slope isequalto , where b = 7 and b2

= 49. Using the values of h, k , a, and b, the equation for


30.center (0, 5); asymptotes y = 5,

conjugate axis length of 12 units

SOLUTION:

The center is (0, 5). Therefore, h = 0 and k = 5. The conjugate axis has a length of 2b units. So, 12

= 2b, b = 6, and b2 = 36. The slopes of the



positive slope isequalto , where a =

, a2 = 19, b = 6, and b

2 = 36.



31.vertices (0, 3), (4, 3); conjugate axis length of 12 units



=1.

The center is located at the midpoint of the

vertices, or (2, 3). So, h = 2 and k = 3. Because the vertices are 4 units apart, 2a = 4, a =

2, and a2 = 4. The length of the conjugate axis is 12

units, so 2b = 12, b = 6, and b2 = 36.






=1.

The center is located at the midpoint of the vertices, or (2, 4). So, h = 2 and k = 4. Because the

vertices are 12 units apart, 2a = 12, a = 6, and a2 =

36. The length of the conjugate axis is 16 units, so

2b = 16, b = 8, and b2 = 64.



33.ARCHITECTUREThefloorplanforanofficebuilding is shown below. a. Write an equation that could model the curved sides of the building. b. Each unit on the coordinate plane represents 15 feet. What is the narrowest width of the building?

SOLUTION:a. From the graph, you can see that the transverse axis is vertical, so the standard form of the equation

is =1.


vertices are 6 units apart, 2a = 6, a = 3, and a2 = 9.

The length of the conjugate axis is 2b = 10. So, b =

5 and b2 = 25.



b. The length of the narrowest width of the building is 6 units. Since each units represents 15 feet, the length of the narrowest width is (15)(6) or 90 feet.

Determine the eccentricity of the hyperbola given by each equation.

34. = 1

SOLUTION:First, find c.

Use the eccentricity equation and the values of c and a to determine the eccentricity of the hyperbola.

c2 = a

2 + b

2

c2 = 10 + 13

c =

e =

=

1.52

35. = 1



c2 = a

2 + b

2

c2 = 24 + 15

c =

e =

=

1.27

36. = 1



c2 = a

2 + b

2

c2 = 38 + 5

c =

e =

=

1.06

37. = 1



c2 = a

2 + b

2

c2 = 32 + 25

c =

e =

=

1.33

38. = 1



c2 = a

2 + b

2

c2 = 23 + 72

c =

e =

=

2.03

39. = 1



c2 = a

2 + b

2

c2 = 16 + 29

c =

e =

=

1.68


40.11x2 2y2 110x + 24y = 181


Next, find c.

Use the eccentricity equation and the values of a and c to determine the eccentricity of the hyperbola.

c2 = a2 + b2

c2 = 2 + 11

c =

e =

=

2.55

41.4x2 + 3y2 + 72x 18y = 321


Next, find c.


c2 = a

2 + b

2

c2 = 8 + 6

c =

e =

=

1.32

42.3x2 2y2 + 12x 12y = 42


Next, find c.


c2 = a2 + b2

c2 = 12 + 18

c =

e =

=

1.58

43.x2 + 7y2 + 24x + 70y = 24


Next, find c.


c2 = a

2 + b

2

c2 = 1 + 7

c =

e =

=

2.83

Use the discriminant to identify each conic section.

44.14y + y2 = 4x 97

SOLUTION:

First, write the equation in the general form Ax2 +

Bxy + Cy2 + Dx + Ey + F = 0.

Next, find the discriminant.

The discriminant is 0, so the conic is a parabola.

14y + y2 = 4x 97

y2 4x + 14y +97 = 0

B2 4AC = 02 4(0)(1)

= 0

45.18x 3x2 + 4 = 8y2 + 32y

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.


The discriminant is greater than 0, so the conic is a hyperbola.

18x 3x2 + 4 = 8y

2 + 32y

3x2 + 8y

2 + 18x 32y +4 = 0

B2 4AC = 0

2 4(3)(8)

= 96

46.14 +4y + 2x2 = 12x y2

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.


The discriminant is less than 0, so the conic must beeither a circle or an ellipse. Because AC, the conic is an ellipse.

14 + 4y + 2x2 = 12x y2

2x2 + y2 + 12x + 4y + 14 = 0

B2 4AC = 02 4(2)(1)

= 8

47.12y 76 x2 = 16x

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



12y 76 x2 = 16x

x2 16x + 12y 76 = 0

B2 4AC = 0

2 4(1)(0)

= 0

48.2x + 8y + x2 + y2 = 8

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.


The discriminant is less than 0, so the conic must beeither a circle or an ellipse. Because A = C, the conic is a circle.

2x + 8y + x2 + y

2 = 8

x2 + y2 + 2x + 8y 8 = 0

B2 4AC = 02 4(1)(1)

= 4

49.5y2 6x + 3x2 50y = 3x2 113

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.


The discriminant is less than 0, so the conic must be

either a circle or an ellipse. Because A C, the conic is an ellipse.

5y2 6x + 3x

2 50y = 3x2 113

6x2 + 5y

2 6x 50y + 113 = 0

B2 4AC = 0

2 4(6)(5)

= 120

50.x2 + y2 + 8x 6y + 9 = 0

SOLUTION:

A = 1, B = 0, and C = 1. Find the discriminant.


B2 4AC = 02 4(1)(1)

= 4

51.56y + 5x2 = 211 + 4y2 + 10x

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.


The discriminant is greater than 0, so the conic must be a hyperbola.

56y + 5x2 = 211 + 4y

2 + 10x

5x2 4y

2 10x 56y 211 = 0

B2 4AC = 0

2 4(5)( 4)

= 80

52.8x + 16 = 8y + 24 x2

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



8x + 16 = 8y + 24 x2

x2 8x 8y 8 = 0

B2 4AC = 02 4(1)(0)

= 0

53.x2 4x = y2 + 12y 31

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



x2 4x = y

2 + 12y 31

x2 + y

2 4x 12y + 31 = 0

B2 4AC = 0

2 4(1)(1)

= 4

54.PHYSICS An hyperbola occurs naturally when two nearly identical glass plates in contact on one edge and separated by about 5 millimeters at the otheredgearedippedinathickliquid.Theliquidwill rise by capillarity to form a hyperbola caused by the surface tension. Find a model for the hyperbola if the conjugate axis is 50 centimeters and the transverse axis is 30 centimeters.

SOLUTION:The length of the conjugate axis is 2b. So, 2b = 50,

b = 25, and b2 = 625. The length of the transverse

axis is 2a. So, 2a = 30, b = 15, and b2 = 225.

Since the orientation of the hyperbola is not

specified, either =1 or

=1 could be used to model the hyperbola.

55.AVIATIONTheFederalAviationAdministrationperforms flight trials to test new technology in aircraft. When one of the test aircraft collected its data, it was 18 kilometers farther from Airport B than Airport A. The two airports are 72 kilometers apart along the same highway, with Airport B due south of Airport A. a. Write an equation for the hyperbola centered at the origin on which the aircraft was located when the data were collected. b. Graph the equation, indicating on which branch of the hyperbola the plane was located. c. When the data were collected, the plane was 40 miles from the highway. Find the coordinates of theplane.

SOLUTION:a. The common difference is 18 kilometers and the absolute value of the difference of the distances from any point on a hyperbola to the foci is 2a, so

2a = 18, a = 9, and a2 = 81. The two airports are

the foci of the hyperbola and are 72 kilometers

apart, so 2c = 72, c = 36, and c2 = 1296. c

2 = a

2 +

b2, so b

2 = 1296 81 or 1215. Airport B is due

south of airport A, so the transverse axis is vertical

and a2term goes with the y2term. The equation

for the hyperbola is =1.

b.

The plane is on the top branch because it is closer to airport A. c. Because the highway is the transverse axis and the plane is 40 kilometers from the highway, x = 40.Substitute 40 for x in the equation.

=1

1.317 = 1

y2 = 187.67

y 13.7 The coordinates of the plane are (40, 13.7)

56.ASTRONOMY Whileeachoftheplanetsinthesolar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.

The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a. 3x2 18x 580850 = 4.84y2 38.72y

b. 360x 8y = y2 1096

c. 24.88y + x2 = 6x 3.11y2 + 412341

SOLUTION:

a. First, write the equation in the general form Ax2

+ Bxy + Cy2 + Dx + Ey + F = 0.

3x2 18x 580850=4.84y

2 38.72y

3x2 4.84y2 18x + 38.72y 580850 = 0

A = 3 and C = 4.84. Find the discriminant.


b. First, write the equation in the general form Ax2

+ Bxy + Cy2 + Dx + Ey + F = 0.

C = 1. Find the discriminant.


c. First, write the equation in the general form Ax2

+ Bxy + Cy2 + Dx + Ey + F = 0.

24.88y + x2

= 6x 3.11y2 + 412341

x2 + 3.11y2 6x 24.88y 412341 = 0




B2 4AC = 0

2 4(3)(4.84)

= 58.08

360x 8y = y2 1096

y2 360x 8y + 1096 = 0

B2 4AC = 0

2 4(0)(1)

= 0

B2 4AC = 0

2 4(1)(3.11)

= 12.44

Derive the general form of the equation for a hyperbola with each of the following characteristics.

57.vertical transverse axis centered at the origin

SOLUTION:From the definition of a hyperbola, you know that the absolute value of the difference of distances from any point P(x, y) on a hyperbola to the foci is

constant, so |PF1 PF2| = 2a. Since you want to

determine the equation for a hyperbola with a vertical transverse axis centered at the origin, use the distance formula d =

andletF1 = (0, c)

and F2 = (0, c).

Use the definition of a hyperbola.

Isolate one radical and then square both sides.

Square both sides again. Simplify, and then

substitute for c2.


1. = 1





2. = 1

SOLUTION:





3. = 1






4. = 1

SOLUTION:





5. = 1






6. =1






7. = 1





8. = 1

SOLUTION:





9.3x2 2y2 = 12







10.3y2 5x2 = 15








photo on Page 449.


0, a = or15,b = or9,andc =





12. = 1





13. = 1






14. = 1





15. = 1





16. = 1






17. = 1





18.x2 4y2 6x 8y = 27






19.x2 + 3y2 4x + 6y = 28



2, k = 1, a = or3,b = orabout5.2,



20.13x2 2y2 + 208x + 16y = 748






21.5x2 + 2y2 70x 8y = 287



7, k = 2, a = or5,b = orabout3.2,




=1, where the



30, a = or30,b = or40,andc =








=1.







a2 = c2 b2

a2 = 82 72

a2 = 64 49

a2 = 15

a =

24.vertices (7, 5), (5, 5); foci (11, 5), (9, 5)



=1.






b2 = c

2 a

2

b2 = 10

2 6

2

b2 = 100 36

b2 = 64

b = 8




=1.








a2 = c2 b2

a2 = 62 32

a2 = 36 9

a2 = 27

a =




=1.








27.vertices (3, 12), (3, 4); foci (3, 15), (3, 1)



=1.









b2 = c2 a2

b2 = 72 42

b2 = 49 16

b2 = 33

b =




=1.





12, and a2 = 144.





SOLUTION:










SOLUTION:






, a2 = 19, b = 6, and b

2 = 36.






=1.




units, so 2b = 12, b = 6, and b2 = 36.






=1.




2b = 16, b = 8, and b2 = 64.





is =1.




5 and b2 = 25.





34. = 1



c2 = a

2 + b

2

c2 = 10 + 13

c =

e =

=

1.52

35. = 1



c2 = a

2 + b

2

c2 = 24 + 15

c =

e =

=

1.27

36. = 1



c2 = a

2 + b

2

c2 = 38 + 5

c =

e =

=

1.06

37. = 1



c2 = a

2 + b

2

c2 = 32 + 25

c =

e =

=

1.33

38. = 1



c2 = a

2 + b

2

c2 = 23 + 72

c =

e =

=

2.03

39. = 1



c2 = a

2 + b

2

c2 = 16 + 29

c =

e =

=

1.68


40.11x2 2y2 110x + 24y = 181


Next, find c.


c2 = a2 + b2

c2 = 2 + 11

c =

e =

=

2.55

41.4x2 + 3y2 + 72x 18y = 321


Next, find c.


c2 = a

2 + b

2

c2 = 8 + 6

c =

e =

=

1.32

42.3x2 2y2 + 12x 12y = 42


Next, find c.


c2 = a2 + b2

c2 = 12 + 18

c =

e =

=

1.58

43.x2 + 7y2 + 24x + 70y = 24


Next, find c.


c2 = a

2 + b

2

c2 = 1 + 7

c =

e =

=

2.83


44.14y + y2 = 4x 97

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



14y + y2 = 4x 97

y2 4x + 14y +97 = 0

B2 4AC = 02 4(0)(1)

= 0

45.18x 3x2 + 4 = 8y2 + 32y

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



18x 3x2 + 4 = 8y

2 + 32y

3x2 + 8y

2 + 18x 32y +4 = 0

B2 4AC = 0

2 4(3)(8)

= 96

46.14 +4y + 2x2 = 12x y2

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



14 + 4y + 2x2 = 12x y2

2x2 + y2 + 12x + 4y + 14 = 0

B2 4AC = 02 4(2)(1)

= 8

47.12y 76 x2 = 16x

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



12y 76 x2 = 16x

x2 16x + 12y 76 = 0

B2 4AC = 0

2 4(1)(0)

= 0

48.2x + 8y + x2 + y2 = 8

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



2x + 8y + x2 + y

2 = 8

x2 + y2 + 2x + 8y 8 = 0

B2 4AC = 02 4(1)(1)

= 4

49.5y2 6x + 3x2 50y = 3x2 113

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.




5y2 6x + 3x

2 50y = 3x2 113

6x2 + 5y

2 6x 50y + 113 = 0

B2 4AC = 0

2 4(6)(5)

= 120

50.x2 + y2 + 8x 6y + 9 = 0

SOLUTION:



B2 4AC = 02 4(1)(1)

= 4

51.56y + 5x2 = 211 + 4y2 + 10x

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



56y + 5x2 = 211 + 4y

2 + 10x

5x2 4y

2 10x 56y 211 = 0

B2 4AC = 0

2 4(5)( 4)

= 80

52.8x + 16 = 8y + 24 x2

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



8x + 16 = 8y + 24 x2

x2 8x 8y 8 = 0

B2 4AC = 02 4(1)(0)

= 0

53.x2 4x = y2 + 12y 31

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



x2 4x = y

2 + 12y 31

x2 + y

2 4x 12y + 31 = 0

B2 4AC = 0

2 4(1)(1)

= 4








55.AVIATIONTheFederalAviationAdministrationperforms flight trials to test new technology in aircraft. When one of the test aircraft collected its data, it was 18 kilometers farther from Airport B than Airport A. The two airports are 72 kilometers apart along the same highway, with Airport B due south of Airport A. a. Write an equation for the hyperbola centered at the origin on which the aircraft was located when the data were collected. b. Graph the equation, indicating on which branch of the hyperbola the plane was located. c. When the data were collected, the plane was 40 miles from the highway. Find the coordinates of theplane.

SOLUTION:a. The common difference is 18 kilometers and the absolute value of the difference of the distances from any point on a hyperbola to the foci is 2a, so

2a = 18, a = 9, and a2 = 81. The two airports are

the foci of the hyperbola and are 72 kilometers

apart, so 2c = 72, c = 36, and c2 = 1296. c

2 = a

2 +

b2, so b

2 = 1296 81 or 1215. Airport B is due

south of airport A, so the transverse axis is vertical

and a2term goes with the y2term. The equation

for the hyperbola is =1.

b.

The plane is on the top branch because it is closer to airport A. c. Because the highway is the transverse axis and the plane is 40 kilometers from the highway, x = 40.Substitute 40 for x in the equation.

=1

1.317 = 1

y2 = 187.67

y 13.7 The coordinates of the plane are (40, 13.7)

56.ASTRONOMY Whileeachoftheplanetsinthesolar system move around the Sun in elliptical orbits, comets may have elliptical, parabolic, or hyperbolic orbits where the center of the sun is a focus.

The paths of three comets are given below, where the values of x and y are measured in gigameters. Use the discriminant to identify each conic.

a. 3x2 18x 580850 = 4.84y2 38.72y

b. 360x 8y = y2 1096

c. 24.88y + x2 = 6x 3.11y2 + 412341

SOLUTION:

a. First, write the equation in the general form Ax2

+ Bxy + Cy2 + Dx + Ey + F = 0.

3x2 18x 580850=4.84y

2 38.72y

3x2 4.84y2 18x + 38.72y 580850 = 0



b. First, write the equation in the general form Ax2

+ Bxy + Cy2 + Dx + Ey + F = 0.

C = 1. Find the discriminant.


c. First, write the equation in the general form Ax2

+ Bxy + Cy2 + Dx + Ey + F = 0.

24.88y + x2

= 6x 3.11y2 + 412341

x2 + 3.11y2 6x 24.88y 412341 = 0




B2 4AC = 0

2 4(3)(4.84)

= 58.08

360x 8y = y2 1096

y2 360x 8y + 1096 = 0

B2 4AC = 0

2 4(0)(1)

= 0

B2 4AC = 0

2 4(1)(3.11)

= 12.44

Derive the general form of the equation for a hyperbola with each of the following characteristics.

57.vertical transverse axis centered at the origin

SOLUTION:From the definition of a hyperbola, you know that the absolute value of the difference of distances from any point P(x, y) on a hyperbola to the foci is

constant, so |PF1 PF2| = 2a. Since you want to

determine the equation for a hyperbola with a vertical transverse axis centered at the origin, use the distance formula d =

andletF1 = (0, c)

and F2 = (0, c).

Use the definition of a hyperbola.

Isolate one radical and then square both sides.

Square both sides again. Simplify, and then

substitute for c2.

eSolutions Manual - Powered by Cognero Page 1

7-3 Hyperbolas


1. = 1





2. = 1

SOLUTION:





3. = 1






4. = 1

SOLUTION:





5. = 1






6. =1






7. = 1





8. = 1

SOLUTION:





9.3x2 2y2 = 12







10.3y2 5x2 = 15








photo on Page 449.


0, a = or15,b = or9,andc =





12. = 1





13. = 1






14. = 1





15. = 1





16. = 1






17. = 1





18.x2 4y2 6x 8y = 27






19.x2 + 3y2 4x + 6y = 28



2, k = 1, a = or3,b = orabout5.2,



20.13x2 2y2 + 208x + 16y = 748






21.5x2 + 2y2 70x 8y = 287



7, k = 2, a = or5,b = orabout3.2,




=1, where the



30, a = or30,b = or40,andc =








=1.







a2 = c2 b2

a2 = 82 72

a2 = 64 49

a2 = 15

a =

24.vertices (7, 5), (5, 5); foci (11, 5), (9, 5)



=1.






b2 = c

2 a

2

b2 = 10

2 6

2

b2 = 100 36

b2 = 64

b = 8




=1.








a2 = c2 b2

a2 = 62 32

a2 = 36 9

a2 = 27

a =




=1.








27.vertices (3, 12), (3, 4); foci (3, 15), (3, 1)



=1.









b2 = c2 a2

b2 = 72 42

b2 = 49 16

b2 = 33

b =




=1.





12, and a2 = 144.





SOLUTION:










SOLUTION:






, a2 = 19, b = 6, and b

2 = 36.






=1.




units, so 2b = 12, b = 6, and b2 = 36.






=1.




2b = 16, b = 8, and b2 = 64.





is =1.




5 and b2 = 25.





34. = 1



c2 = a

2 + b

2

c2 = 10 + 13

c =

e =

=

1.52

35. = 1



c2 = a

2 + b

2

c2 = 24 + 15

c =

e =

=

1.27

36. = 1



c2 = a

2 + b

2

c2 = 38 + 5

c =

e =

=

1.06

37. = 1



c2 = a

2 + b

2

c2 = 32 + 25

c =

e =

=

1.33

38. = 1



c2 = a

2 + b

2

c2 = 23 + 72

c =

e =

=

2.03

39. = 1



c2 = a

2 + b

2

c2 = 16 + 29

c =

e =

=

1.68


40.11x2 2y2 110x + 24y = 181


Next, find c.


c2 = a2 + b2

c2 = 2 + 11

c =

e =

=

2.55

41.4x2 + 3y2 + 72x 18y = 321


Next, find c.


c2 = a

2 + b

2

c2 = 8 + 6

c =

e =

=

1.32

42.3x2 2y2 + 12x 12y = 42


Next, find c.


c2 = a2 + b2

c2 = 12 + 18

c =

e =

=

1.58

43.x2 + 7y2 + 24x + 70y = 24


Next, find c.


c2 = a

2 + b

2

c2 = 1 + 7

c =

e =

=

2.83


44.14y + y2 = 4x 97

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



14y + y2 = 4x 97

y2 4x + 14y +97 = 0

B2 4AC = 02 4(0)(1)

= 0

45.18x 3x2 + 4 = 8y2 + 32y

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



18x 3x2 + 4 = 8y

2 + 32y

3x2 + 8y

2 + 18x 32y +4 = 0

B2 4AC = 0

2 4(3)(8)

= 96

46.14 +4y + 2x2 = 12x y2

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



14 + 4y + 2x2 = 12x y2

2x2 + y2 + 12x + 4y + 14 = 0

B2 4AC = 02 4(2)(1)

= 8

47.12y 76 x2 = 16x

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



12y 76 x2 = 16x

x2 16x + 12y 76 = 0

B2 4AC = 0

2 4(1)(0)

= 0

48.2x + 8y + x2 + y2 = 8

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



2x + 8y + x2 + y

2 = 8

x2 + y2 + 2x + 8y 8 = 0

B2 4AC = 02 4(1)(1)

= 4

49.5y2 6x + 3x2 50y = 3x2 113

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.




5y2 6x + 3x

2 50y = 3x2 113

6x2 + 5y

2 6x 50y + 113 = 0

B2 4AC = 0

2 4(6)(5)

= 120

50.x2 + y2 + 8x 6y + 9 = 0

SOLUTION:



B2 4AC = 02 4(1)(1)

= 4

51.56y + 5x2 = 211 + 4y2 + 10x

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



56y + 5x2 = 211 + 4y

2 + 10x

5x2 4y

2 10x 56y 211 = 0

B2 4AC = 0

2 4(5)( 4)

= 80

52.8x + 16 = 8y + 24 x2

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



8x + 16 = 8y + 24 x2

x2 8x 8y 8 = 0

B2 4AC = 02 4(1)(0)

= 0

53.x2 4x = y2 + 12y 31

SOLUTION:


Bxy + Cy2 + Dx + Ey + F = 0.



x2 4x = y

2 + 12y 31

x2 + y

2 4x 12y + 31 = 0

B2 4AC = 0

2 4(1)(1)

= 4








55.AVIATIONTheFederalAviationAdministrationperforms flight trials to test new technology in aircraft. When one of the test aircraft collected its data, it was 18 kilometers farther from Airport B than Airport A. The two airports are 72 kilometers apart along the same highway, with Airport B due south of Airport A. a. Write an equation for the hyperbola centered at the origin on which the aircraft was located when the data were collected. b. Graph the equation, indicating on which branch of the hyperbola the plane was located. c. When the data were collected, the plane was 40 miles from the highway. Find the coordinates of t

eSolutions Manual - Powered by Cognero Page 1 · foci: (h, k c) = (0, 2.83) and ( 0, í2.83 ) Graph...

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Transcript of eSolutions Manual - Powered by Cognero Page 1 · foci: (h, k c) = (0, 2.83) and ( 0, í2.83 ) Graph...