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10.6 – Translating Conic Sections

10.6 – Translating Conic Sections. Translating Conics means that we move them from the initial position with an origin at (0, 0) (the parent graph) to a new position. Equations of Conic Sections. Practice.

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10.6 – Translating Conic Sections

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  1. 10.6 – Translating Conic Sections

  2. Translating Conics means that we move them from the initial position with an origin at (0, 0) (the parent graph) to a new position

  3. Equations of Conic Sections

  4. Practice Name the parent function for the equations in Exercises 1–4. Describe each equation as a translation of the parent function. 1.y = x2 + 4 2.y = (x – 3)2 – 2 3.y – 1 = x24.y = (x + 5)2 + 6 Rewrite each equation in vertex form. Hint: you may need to complete the square! 5.y = x2 – 6x + 1 6.y = x2 + 10x – 7 7.y = 2x2 + 8x + 5 8.y = 4x2 – 12x + 3

  5. Solutions 1.y = x2 + 4 2.y = (x – 3)2 – 2 parent function: y = x2; parent function: y = x2; translated translated 4 units up 3 units right and 2 units down 3.y – 1 = x2, or y = x2 + 1 4.y = (x + 5)2 + 6 parent function: y = x2; parent function: y = x2; translated 1 unit up translated 5 units left and 6 units up 5.y = x2 – 6x + 1; c = – 26.y = x2 + 10x – 7; c = 2 = (–3)2 = 9 = 52 = 25 y = (x2 – 6x + 9) + 1 – 9 y = (x2 + 10x + 25) – 7 – 25 y = (x – 3)2 – 8 y = (x + 5)2 – 32 6 2 10 2

  6. 7.y = 2x2 + 8x + 5 8.y = 4x2 – 12x + 3 y = 2(x2 + 4x) + 5; c = 2y = 4(x2 – 3x) + 3; c = 2 = 22 = 4 = y = 2(x2 + 4x + 4) + 5 – 2(4) y = 4 x2 – 3x + + 3 – 4 y = 2(x + 2)2 + 5 – 8 y = 4 2 + 3 – 9 y = 2(x + 2)2 – 3 y = 4 2 – 6 3 2 – 9 4 9 4 9 4 4 2 3 2 3 2 x – x – Solutions (continued)

  7. The major axis is vertical, so the equation has the form + = 1. (y – k)2 a2 (x – h)2 b2 Substitute –2 for h and 4 for k. + = 1. (y – 4)2 52 (x – (–2))2 42 (x + 2)2 16 (y – 4)2 25 + = 1. The equation of the ellipse is Example Write an equation of an ellipse with center (–2, 4), a vertical major axis of length 10, and minor axis of length 8. The length of the major axis is 2a. So 2a = 10 and a = 5. The length of the minor axis is 2b. So 2b = 8 and b = 4. Since the center is (–2, 4), h = –2 and k = 4.

  8. Check: Solve the equation for y and graph both equations. 25(x + 2)2 + 16(y – 4)2 = 400 16(y – 4)2 = 400 – 25(x + 2)2 (y – 4)2 = (400 – 25(x + 2)2) 1 16 1 16 (x + 2)2 16 (y – 4)2 25 + = 1. y – 4 = ± (400 – 25(x + 2)2) 1 4 y = 4 ± 400 – 25(x + 2)2 Example (continued)

  9. Draw a sketch. The center is the midpoint of the line joining the vertices. Its coordinates are (1, 2). The distance between the vertices is 2a, and the distance between the foci is 2c. 2a = 4, so a = 2; 2c = 8, so c = 4. Example Write an equation of a hyperbola with vertices (–1, 2) and (3, 2), and foci (–3, 2) and (5, 2). Find b2 using the Pythagorean Theorem. c2 = a2 + b2 16 = 4 + b2 b2 = 12

  10. The transverse axis is horizontal, so the equation has the form – = 1. (x – h)2 a2 (x – 1)2 4 (y – 2)2 12 (y – k)2 b2 – = 1. The equation of the hyperbola is Example (continued)

  11. Since 2c = 80, c = 40. The center of the hyperbola is at (40, 0). Find a by calculating the difference in the distances from the vertex at (a + 40, 0) to the two foci. 30 = (a + 40) – (80 – (a + 40)) = 2a 15 = a Example Use the information from Example 3. Find the equation of the hyperbola if the transmitters are 80 mi apart located at (0, 0) and (80, 0), and all points on the hyperbola are 30 mi closer to one transmitter than the other.

  12. y2 1375 y2 1375 (x – 40)2 152 (x – 40)2 225 – = 1 The equation of the hyperbola is or – = 1. (continued) Find b2. c2 = a2 + b2 (40)2 = (15)2 + b2 1600 = 225 + b2 b2 = 1375

  13. 9x2 – 4y2 + 18x = 27 9x2 + 18x – 4y2 = 27 Group the x- and y- terms. 9(x2 + 2x + ) – 4y2 = 27 Complete the square. 9(x2 + 2x + 1) – 4y2 = 27 + 9(12) Add (9)(12) to each side. 9(x2 + 2x + 1) – 4y2 = 27 + 9Simplify 9(x + 1)2 – 4y2 = 36 Write the trinomials as binomials squared. Example Identify the conic section with equation 9x2 – 4y2 + 18x = 27. If it is a parabola, give the vertex. If it is a circle, give the center and the radius. If it is an ellipse or a hyperbola, give the center and foci. Sketch the graph. Complete the square for the x- and y-terms to write the equation in standard form.

  14. – Divide each side by 36. = 1 Simplify. 9(x + 1)2 36 4y2 36 (x + 1)2 4 – = 1 y2 9 c2 = a2 + b2 = 4 + 9 = 13 c = 13 Translating Conic Sections (continued) The equation represents a hyperbola. The center is (–1, 0). The transverse axis is horizontal. Since a2 = 4, a = 2, b2 = 9, so b = 3.

  15. The distance from the center of the hyperbola to the foci is 13. Since the hyperbola is centered at (–1, 0), and the transverse axis is horizontal, the foci are located 13 to the left and right of the center. The foci are at (–1 + 13, 0) and (–1 – 13, 0). Translating Conic Sections (continued)

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