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 that A = 4, B = 0, and C = 1, so the value of the discriminant is: B 2 – 4AC = 0 2 – 4(4)(1) = –16 Because B 2 – 4AC < 0 and A = C, the conic is an ellipse. To 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

EXAMPLE 6 Classify a conic 4(x 2 – 2x + 1) + y 2 = 8 + 4(1) 4(x – 1) 2 + y 2 = 12 (x – 1) y 2 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 16x 2 – 9y 2 – 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. B 2 – 4AC = 0 2 – 4(16)(–9) = 576 Because B 2 – 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. 16x 2 – 9y 2 – 96x + 36y – 36 = 0 (16x 2 – 96x) – (9y 2 – 36y) = 36 16(x 2 – 6x + ? ) – 9(y 2 – 4y + ? ) = ( ? ) – 9( ? ) 16(x 2 – 6x + 9) – 9(y 2 – 4y + 4) = (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 Classify the conic given by x 2 + y 2 – 2x + 4y + 1 = 0. Then graph the equation. (x – 1) 2 +( y + 2) 2 = 4 ANSWER

GUIDED PRACTICE for Examples 6 and Classify the conic given by 2x 2 + y 2 – 4x – 4 = 0. Then graph the equation. (x – 1) 2 3 y2y2 6 + = 1 ANSWER Circle

GUIDED PRACTICE for Examples 6 and Classify the conic given by y 2 – 4y 2 – 2x + 6 = 0. Then graph the equation. (y – 2) 2 = 2(x –1) ANSWER Parabola

GUIDED PRACTICE for Examples 6 and Classify the conic given by 4x 2 – y 2 – 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 Astronomy An asteroid’s path is modeled by 4x y 2 – 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 – y2y2 4 = 1 ANSWER Hyperbola