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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( ? )

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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.

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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.

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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.

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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

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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).

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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.

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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 .

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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

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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 .

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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.

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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 .

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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

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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 .

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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.

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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 .

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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

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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

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