4( x 2 – 2 x + ? ) + y 2 = 8 + 4( ? )
DESCRIPTION
Note that A = 4, B = 0, and C = 1 , so the value of the discriminant is:. B 2 – 4 AC = 0 2 – 4(4)(1) = –16. Because B 2 – 4 AC < 0 and A = C, the conic is an ellipse. To graph the ellipse, first complete the square in x. EXAMPLE 6. Classify a conic. - PowerPoint PPT PresentationTRANSCRIPT
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.
(x – 1)2 +( y + 2)2 = 4
ANSWER
GUIDED PRACTICE for Examples 6 and 7
11. Classify the conic given by 2x2 + y2 – 4x – 4 = 0. Then graph the equation.
(x – 1)2
3
y2
6+ = 1
ANSWER
Circle
GUIDED PRACTICE for Examples 6 and 7
12. Classify the conic given by y2 – 4y2 – 2x + 6 = 0. Then graph the equation.
(y – 2)2 = 2(x –1)
ANSWER
Parabola
GUIDED PRACTICE for Examples 6 and 7
13. Classify the conic given by 4x2 – y2 – 16x – 4y – 4 = 0. Then graph the equation.
ANSWER
(x –2)2
4
(y +2)2
16= 1–
Hyperbola
GUIDED PRACTICE for Examples 6 and 7
14. Astronomy
An asteroid’s path is modeled by 4x2 + 6.25y2 – 12x – 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.
4(x – 1.5)2
4–
y2
4= 1
ANSWER
Hyperbola