1019 10.3 Hyperbolas x y 3 –3 0 4 –4 Horizontal transverse axis – = 1 9 (x – 3)2 16x2 – 96x – 9y2 – 72y = 144 16 (y + 4)2 (3, –4) Figure 34 EXAMPLE 4 Writing the Equation of a Hyperbola in Standard Form Write the equation of the hyperbola 16x2 - 96x - 9y2 - 72y = 144 in standard form. Give the center, and determine whether the transverse axis is horizontal or vertical. SOLUTION 16x2 - 96x - 9y2 - 72y = 144 Given equation 161x2 - 6x 2 - 91y2 + 8y 2 = 144 Factor out 16, and factor out -9. Think: c 1 2 1-62d 2 = 9 and c 1 2 182d 2 = 16 Prepare to complete the square for both x and y. 161x2 - 6x + 9 - 92 - 91y2 + 8y + 16 - 162 = 144 Complete the square. 161x2 - 6x + 92 - 144 - 91y2 + 8y + 162 + 144 = 144 Distributive property 161x - 322 - 91y + 422 = 144 Factor. Combine like terms. 1x - 322 9 - 1y + 422 16 = 1 Divide by 144. This equation represents a hyperbola with center 13, -42. The transverse axis is horizontal because the leading term—that is, the positive term—of the hyperbola written in standard form contains the x2-term. See Figure 34. S Now Try Exercise 31. Multiply 161-92 = -144 and -91-162 = 144 Eccentricity If we apply the definition of eccentricity from the previous section to the hyperbola, we obtain the following. e = 2a2 + b2 a = c a Eccentricity of a hyperbola Because c 7a, we have e 71. Thus, every hyperbola has eccentricity greater than 1. Narrow hyperbolas have e near 1, and wide hyperbolas have large e. See Figure 35. x y e = 1.1 x y e = 2 x y e = 5 Figure 35 Eccentricities of hyperbolas EXAMPLE 5 Finding Eccentricity from the Equation of a Hyperbola Find the eccentricity of the hyperbola x2 9 - y2 4 = 1. Round to the nearest tenth. SOLUTION Here, a2 = 9 and thus a = 3. Also, b2 = 4. c2 = a2 + b2 Relationship for hyperbolas c2 = 9 + 4 Let a2 = 9 and b2 = 4. c2 = 13 Add. c = 213 Take the positive square root because c 70. eccentricity e = c a = 213 3 ≈1.2 S Now Try Exercise 39.
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