x 2 – 2x + ? ) + y 2 = 8 + 4( ? ) EXAMPLE 6 Classify a conic Classify the conic given by 4x 2 + y 2 – 8x – 8 = 0. Then graph the equation. SOLUTION Note that A = 4, B = 0, and C = 1, so the value of the discriminant is: B 2 – 4AC = 0 2 – 4(4)(1) = – 16 cause B 2 – 4AC < 0 and A = C, the conic is an ellipse graph the ellipse, first complete the square in x. 4x 2 + y 2 – 8x – 8 = 0 (4x 2 – 8x) + y 2 = 8 4(x 2 – 2x) + y 2 = 8
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4(x 2 – 2x + ? ) + y 2 = 8 + 4( ? ) EXAMPLE 6 Classify a conic Classify the conic given by 4x 2 + y 2 – 8x – 8 = 0. Then graph the equation. SOLUTION Note.
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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.
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