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Graphing Conics and Classifying Shapes

This text provides examples of how to graph and classify conics, including ellipses, circles, parabolas, and hyperbolas. It also demonstrates how to write the equations for conics in standard form and provides step-by-step instructions for graphing the equations.

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Graphing Conics and Classifying Shapes

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  1. Note thatA = 4, B = 0,andC = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(4)(1) = – 16 BecauseB2– 4AC < 0andA = C, the conic is an ellipse. To graph the ellipse, first complete the square in x. EXAMPLE 6 Classify a conic Classify the conic given by4x2 + y2 – 8x – 8 = 0. Then graph the equation. SOLUTION 4x2 + y2 – 8x – 8 = 0 (4x2 – 8x) + y2 = 8 4(x2 – 2x) + y2 = 8 4(x2– 2x + ? ) + y2 = 8 + 4( ? )

  2. (x – 1)2 y2 12 + = 1 3 From the equation, you can see that(h, k) = (1, 0), a = 12 = 2 3 ,andb = 3. Use these facts to draw the ellipse. EXAMPLE 6 Classify a conic 4(x2– 2x + 1) + y2 = 8 + 4(1) 4(x – 1)2 + y2 = 12

  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 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.

  4. EXAMPLE 7 Solve a multi-step problem SOLUTION STEP 1 Identify the shape. The equation is a general second-degree equation withA = 16, B = 0,andC = – 9. Find the value of the discriminant. B2 – 4AC = 02 – 4(16)(– 9) = 576 BecauseB2 – 4AC > 0, the shape of the path is a hyperbola.

  5. (x – 3)2 9 (y –2)216 – = 1 EXAMPLE 7 Solve a multi-step problem STEP 2 Write an equation. To write an equation of the hyperbola, complete the square in both xand ysimultaneously. 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

  6. Graph the equation. From the equation, the transverse axis is horizontal, (h, k) = (3, 2), a = 9 = 3 andb = 16. = 4 EXAMPLE 7 Solve a multi-step problem STEP 3 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. Note thatA = 1, B = 0,andC = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(1)(1) = – 4 BecauseB2 – 4AC < 0andA = C, the conic is an circle. To graph the circle, first complete the square in both x and y simultaneity. for Examples 6 and 7 GUIDED PRACTICE 10. Classify the conic given byx2 + y2 – 2x + 4y + 1 = 0. Then graph the equation. SOLUTION

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

  10. Note thatA = 2, B = 0,andC = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(2)(1) = – 8 BecauseB2 – 4AC < 0andA = C, the conic is an circle. for Examples 6 and 7 GUIDED PRACTICE 11. Classify the conic given by2x2 + y2 – 4x – 4= 0. Then graph the equation. SOLUTION To graph the ellipse , complete the square x .

  11. + y2 = 1 (x – 1)2 6 3 ANSWER From the equation, you can see that(h, k) = (1, 0), a = 3 andb = 6. Use these facts to draw the ellipse. for Examples 6 and 7 GUIDED PRACTICE 2x2 + y2 – 4x – 4= 0 2x2 – 4x +2+ y2 = 6 2(x – 1)2 + y2= 6

  12. Note thatA = 0, B = 0,andC = 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(0)(1) = 0 BecauseB2 – 4AC < 0andA = C, the conic is an circle. for Examples 6 and 7 GUIDED PRACTICE 12. Classify the conic given byy2 – 4y2 – 2x + 6= 0. Then graph the equation. SOLUTION To graph the parabola , complete the square y .

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

  14. Note thatA = 4, B = 0,andC = – 1, so the value of the discriminant is: B2 – 4AC = 02 – 4(4)(– 1) = 16 BecauseB2 – 4AC > 0andA = C, the conic is an circle. for Examples 6 and 7 GUIDED PRACTICE 13. Classify the conic given by4x2 – y2 – 16x –4y – 4= 0. Then graph the equation. SOLUTION To graph the parabola , complete the square in both x and y simultaneity.

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

  16. for Examples 6 and 7 GUIDED PRACTICE 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 1 Identify the shape. The equation is a general second-degree equation withA = 4, B = 0,andC = 6.25. Find the value of the discriminant. B2 – 4AC = 0 – 4(4)(6.25) = – 100 BecauseB2 – 4AC < 0, the shape of the path is a ellipse .

  17. 4(x2–3x + ) +6.25y2 – 16 – 9 = 0 4(x – )2+ 6.25y2 = 25 (x – )2 – = 1 (x – 1.5)2 3 3 3 (y)225 4 2 2 2 4 y2 + = 1 25 for Examples 6 and 7 GUIDED PRACTICE STEP 2 4x2+6.25y2– 12x – 16 = 0 (4x2–12x) –(6.25y2– 16) = 0

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

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