1. 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
4(x2 – 2x + ? ) + y2 = 8 + 4( ? )

2. 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 = , and b = 3. Use these facts to draw the ellipse.

3. 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 x2 – 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.

4. 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.

5. 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 + ? ) = ( ? ) – 9( ? )
16(x2 – 6x + 9) – 9(y2 – 4y + 4) = (9) – 9(4)
16(x – 3)2 – 9(y – 2)2 = 144
(x – 3)2 9 –
(y –2)
= 1

6. 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 = = 3 and b = = 4
The vertices are at (3 + a, 2), or (6, 2) and (0, 2).

7. 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.

8. GUIDED PRACTICE
for Examples 6 and 7
10.
Classify the conic given by x2 + y2 – 2x + 4y + 1 = 0. Then graph the equation.
SOLUTION
Note that A = 1, B = 0, and C = 1, so the value of the discriminant is:
B2 – 4AC = 02 – 4(1)(1) = – 4
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 .

9. 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

10. GUIDED PRACTICE
for Examples 6 and 7
11.
Classify the conic given by 2x2 + y2 – 4x – 4 = 0. Then graph the equation.
SOLUTION
Note that A = 2, B = 0, and C = 1, so the value of the discriminant is:
B2 – 4AC = 02 – 4(2)(1) = – 8
Because B2 – 4AC < 0 and A = C, the conic is an circle.
To graph the ellipse , complete the square x .

11. 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.

12. GUIDED PRACTICE
for Examples 6 and 7
12.
Classify the conic given by y2 – 4y2 – 2x + 6 = 0. Then graph the equation.
SOLUTION
Note that A = 0, B = 0, and C = 1, so the value of the discriminant is:
B2 – 4AC = 02 – 4(0)(1) = 0
Because B2 – 4AC < 0 and A = C, the conic is an circle.
To graph the parabola , complete the square y .

13. 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

14. GUIDED PRACTICE
for Examples 6 and 7
13.
Classify the conic given by 4x2 – y2 – 16x – 4y – 4 = 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 circle.
To graph the parabola , complete the square in both x and y simultaneity .

15. GUIDED PRACTICE
for Examples 6 and 7
4x2 – y2 – 16x – 4y – y = 0
(4x2 – 16x + 16) –
(y2 + 4y +4) = 16
4(x2 – 4x + 4) –
(y2 + 4y + 4) = 16
4(x – 2)2 –
(y + 2)2
= 16
(x –2)2 4
(y +2)2 16
= 1 –
ANSWER
From the equation, you can see that (h, k) = (2, – 2), a = 2 and b = 4 . Use these facts to draw the parabola.

16. GUIDED PRACTICE
for Examples 6 and 7
14.
Astronomy
An asteroid’s path is modeled by 4x y2 – 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 1
Identify the shape. The equation is a general second-degree equation with A = 4, B = 0, and C = 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 .

17. GUIDED PRACTICE
for Examples 6 and 7
STEP 2
4x y2 – 12x – 16 = 0
(4x2 – 12x) – (6.25y2 – 16) = 0
4(x2 – 3x ) +6.25y2 – 16 – 9 = 0 3 2
4(x – ) y2 = 25 3 2
(x – )2 4 –
(y)
= 1 3 2
(x – 1.5)2 4 + y2 25
= 1

18. GUIDED PRACTICE
for Examples 6 and 7
STEP 3
ANSWER
From the equation, you can see that (h, k) = (1.5, 0), a = and b = 2.
6.25