Translating Conic Sections ALGEBRA 2 LESSON 10-6 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. The major axis is vertical, so the equation has the form (x – h)2 b2 (y – k)2 a2 + = 1. (x – (–2))2 42 (y – 4)2 52 + = 1. Substitute –2 for h and 4 for k. The equation of the ellipse is + = 1. (x + 2)2 16 (y – 4)2 25 10-6

Translating Conic Sections ALGEBRA 2 LESSON 10-6 (continued) (y – 4)2 = (400 – 25(x + 2)2) 1 16 Check: Solve the equation for y and graph both equations. + = 1. (x + 2)2 (y – 4)2 25 25(x + 2)2 + 16(y – 4)2 = 400 16(y – 4)2 = 400 – 25(x + 2)2 y – 4 = ± (400 – 25(x + 2)2) y = 4 ± 400 – 25(x + 2)2 4 10-6

Translating Conic Sections ALGEBRA 2 LESSON 10-6 Write an equation of a hyperbola with vertices (–1, 2) and (3, 2), and foci (–3, 2) and (5, 2). 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. Find b2 using the Pythagorean Theorem. c2 = a2 + b2 16 = 4 + b2 b2 = 12 10-6

Translating Conic Sections ALGEBRA 2 LESSON 10-6 (continued) The transverse axis is horizontal, so the equation has the form (x – h)2 a2 (y – k)2 b2 – = 1. The equation of the hyperbola is (x – 1)2 4 (y – 2)2 12 – = 1. 10-6

Translating Conic Sections ALGEBRA 2 LESSON 10-6 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. 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 10-6

Translating Conic Sections ALGEBRA 2 LESSON 10-6 (continued) Find b2. c2 = a2 + b2 (40)2 = (15)2 + b2 1600 = 225 + b2 b2 = 1375 The equation of the hyperbola is (x – 40)2 152 y 2 1375 – = 1 or 225 = 1. 10-6

Translating Conic Sections ALGEBRA 2 LESSON 10-6 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. 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 + 9 Simplify 9(x + 1)2 – 4y2 = 36 Write the trinomials as binomials squared. 10-6

Translating Conic Sections ALGEBRA 2 LESSON 10-6 (continued) 9(x + 1)2 36 4y 2 – = 1 Divide each side by 36. (x + 1)2 4 y 2 9 Simplify. 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. c2 = a2 + b2 = 4 + 9 = 13 c = 13 10-6

Translating Conic Sections ALGEBRA 2 LESSON 10-6 (continued) 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). 10-6