1020 CHAPTER 10 Analytic Geometry EXAMPLE 6 Writing an Equation of a Hyperbola Write an equation for the hyperbola with e = 2 and foci at 1-9, 52 and 1-3, 52. SOLUTION Because the foci have the same y-coordinate, the line through them, and therefore the hyperbola, is horizontal. The center of the hyperbola is halfway between the two foci at 1-6, 52. The distance from each focus to the center is c = 3, so c2 = 9. Because e = c a, we have a = c e = 3 2 and a2 = 9 4. c2 = a2 + b2 Relationship for hyperbolas 9 = 9 4 + b2 Let c2 = 9 and a2 = 9 4. b2 = 27 4 Solve for b2; 9 - 9 4 = 36 4 - 9 4 = 27 4 The equation of the hyperbola is 1x + 622 9 4 - 1y - 522 27 4 = 1, or 41x + 622 9 - 41y - 522 27 = 1. Simplify complex fractions. S Now Try Exercise 57. 10.3 Exercises CONCEPT PREVIEW Match each equation of a hyperbola in Column I with its description in Column II. The following chart summarizes our discussion of eccentricity in this chapter. Summary of Eccentricity Conic Section Eccentricity e Parabola e = 1 Circle e = 0 Ellipse e = c a and 0 6e 61 Hyperbola e = c a and e 71 II A. center 11, 22; horizontal transverse axis B. center 1-2, -12; vertical transverse axis C. center 1-1, -22; horizontal transverse axis D. center 12, 12; vertical transverse axis I 1. 1x - 122 49 - 1y - 222 64 = 1 2. 1x + 122 64 - 1y + 222 49 = 1 3. 1y - 122 9 - 1x - 222 25 = 1 4. 1y + 122 25 - 1x + 222 9 = 1
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