4(x2 – 2x + ? ) + y2 = 8 + 4( ? )

EXAMPLE 6 Classify a conic

Classify the conic given by 4x2 + y2 – 8x – 8 = 0. Then graph the equation.

SOLUTION

Note that A = 4, B = 0, and C = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(4)(1) = – 16

Because B2– 4AC < 0 and A = C, the conic is an ellipse.To graph the ellipse, first complete the square in x.

4x2 + y2 – 8x – 8 = 0

(4x2 – 8x) + y2 = 8

4(x2 – 2x) + y2 = 8

EXAMPLE 6 Classify a conic

4(x2 – 2x + 1) + y2 = 8 + 4(1)

4(x – 1)2 + y2 = 12

(x – 1)2

3+

y2

12 = 1

From the equation, you can see that (h, k) = (1, 0), a = 12 = 2 3 , and b = 3. Use these facts to draw the ellipse.

EXAMPLE 7 Solve a multi-step problem

Physical Science

In a lab experiment, you record images of a steel ball rolling past a magnet. The equation 16x2 – 9y2 – 96x + 36y – 36 = 0 models the ball’s path.• What is the shape of the path ?

• Write an equation for the path in standard form.

• Graph the equation of the path.

EXAMPLE 7 Solve a multi-step problem

SOLUTION

STEP 1

Identify the shape. The equation is a general second-degree equation with A = 16, B = 0, and C = – 9. Find the value of the discriminant.

B2 – 4AC = 02 – 4(16)(– 9) = 576

Because B2 – 4AC > 0, the shape of the path is a hyperbola.

EXAMPLE 7 Solve a multi-step problem

STEP 2

Write an equation. To write an equation of the hyperbola, complete the square in both x and y simultaneously.

16x2 – 9y2 – 96x + 36y – 36 = 0(16x2 – 96x) – (9y2 – 36y) = 36

16(x2 – 6x + ? ) – 9(y2 – 4y + ? ) = 36 + 16( ? ) – 9( ? )

16(x2 – 6x + 9) – 9(y2 – 4y + 4) = 36 + 16(9) – 9(4)16(x – 3)2 – 9(y – 2)2 = 144

(x – 3)2

9–

(y –2)2 16

= 1

EXAMPLE 7 Solve a multi-step problem

STEP 3

Graph the equation. From the equation, the transverse axis is horizontal, (h, k) = (3, 2),

a = 9 = 3 and b = 16. = 4

The vertices are at (3 + a, 2), or (6, 2) and (0, 2).

EXAMPLE 7 Solve a multi-step problem

STEP 3

Plot the center and vertices. Then draw a rectangle 2a = 6 units wide and 2b = 8 units high centered at (3, 2), draw the asymptotes, and draw the hyperbola.

Notice that the path of the ball is modeled by just the right-hand branch of the hyperbola.

GUIDED PRACTICE for Examples 6 and 7

10. Classify the conic given by x2 + y2 – 2x + 4y + 1 = 0. Then graph the equation.

Note that A = 1, B = 0, and C = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(1)(1) = – 4

SOLUTION

Because B2 – 4AC < 0 and A = C, the conic is an circle.To graph the circle, first complete the square in both x and y simultaneity .

GUIDED PRACTICE for Examples 6 and 7

x2 + y2 – 2x + 4y + 1 = 0

x2 – 2x +1+ y2 + 4y + 4 = 4

(x – 1)2 +( y + 2)2 = 4

From the equation, you can see that (h, k) = (– 1, 2), r = 2 Use these facts to draw the circle.

ANSWER

GUIDED PRACTICE for Examples 6 and 7

11. Classify the conic given by 2x2 + y2 – 4x – 4 = 0. Then graph the equation.

Note that A = 2, B = 0, and C = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(2)(1) = – 8

SOLUTION

To graph the ellipse , complete the square x .

Because B2 – 4AC < 0 and A = C, the conic is an circle.

GUIDED PRACTICE for Examples 6 and 7

2x2 + y2 – 4x – 4 = 0

2x2 – 4x +2+ y2 = 6

2(x – 1)2 + y2 = 6

(x – 1)2

3

y2

6+ = 1

ANSWER

From the equation, you can see that (h, k) = (1, 0), a = 3 and b = 6. Use these facts to draw the ellipse.

GUIDED PRACTICE for Examples 6 and 7

12. Classify the conic given by y2 – 4y2 – 2x + 6 = 0. Then graph the equation.

Note that A = 0, B = 0, and C = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(0)(1) = 0

SOLUTION

To graph the parabola , complete the square y .

Because B2 – 4AC < 0 and A = C, the conic is an circle.

GUIDED PRACTICE for Examples 6 and 7

y2 – 4y2 – 2x + 6 = 0

y2 – 4y + 4 = 2x – 2

(y – 2)2 = 2(x –1)

ANSWER

From the equation, you can see that (h, k) = (1, 2), p = . Use these facts to draw the parabola.1

2

GUIDED PRACTICE for Examples 6 and 7

13. Classify the conic given by 4x2 – y2 – 16x – 4y – 4 = 0. Then graph the equation.

Note that A = 4, B = 0, and C = – 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(4)(– 1) = 16

SOLUTION

To graph the parabola , complete the square in both x and y simultaneity .

Because B2 – 4AC > 0 and A = C, the conic is an circle.

GUIDED PRACTICE for Examples 6 and 7

4x2 – y2 – 16x – 4y – y = 0

ANSWER

From the equation, you can see that (h, k) = (2, – 2), a = 2 and b = 4 . Use these facts to draw the parabola.

(4x2 – 16x + 16) – (y2 + 4y +4) = 16

4(x2 – 4x + 4) – (y2 + 4y + 4) = 16

(x –2)2

4

(y +2)2

16= 1–

4(x – 2)2 – (y + 2)2 = 16

GUIDED PRACTICE for Examples 6 and 7

14. Astronomy

An asteroid’s path is modeled by 4x2 + 6.25y2 – 12y – 16 = 0 where x and y are in astronomical units from the sun. Classify the path and write its equation in standard form. Then graph the equation.

SOLUTION

STEP 1Identify the shape. The equation is a general second-degree equation with A = 4, B = 0, and C = 6.25. Find the value of the discriminant.

B2 – 4AC = 0 – 4(4)(6.25) = – 100

Because B2 – 4AC < 0, the shape of the path is a ellipse .

GUIDED PRACTICE for Examples 6 and 7

4x2 + 6.25y2 – 12x – 16 = 0

(4x2 – 12x) – (6.25y2 – 16) = 0

(x – )2

4–

(y)2 25

= 132

4(x2 – 3x + ) +6.25y2 – 16 – 9 = 032

4(x – )2 + 6.25y2 = 2532

(x – 1.5)2

4+

y2

25= 1

STEP 2

STEP 3

GUIDED PRACTICE for Examples 6 and 7

ANSWER

From the equation, you can see that (h, k) = (1.5, 0), a = and b = 2. 6.25