1018 CHAPTER 10 Analytic Geometry Translated Hyperbolas Like an ellipse, a hyperbola can have its center translated away from the origin. Standard Forms for Hyperbolas Centered at 1h, k2 A hyperbola with center 1h, k2 and either a horizontal or a vertical transverse axis satisfies one of the following equations, where c2 = a2 + b2. 1x −h22 a2 − 1 y −k22 b2 =1 Transverse axis: horizontal; vertices: 1h{a, k2; foci: 1h{c, k2; asymptotes: y = { b a 1x - h2 + k 1y −k22 a2 − 1 x −h22 b2 =1 Transverse axis: vertical; vertices: 1h, k {a2; foci: 1h, k {c2; asymptotes: y = { a b 1x - h2 + k x y 0 a a c c Horizontal transverse axis (h, k) F9 F b b x y 0 a a c c b b Vertical transverse axis (h, k) F9 F NOTE The asymptotes for a hyperbola always pass through the center 1h, k2. By the point-slope form of a line, the equation of any asymptote is y = m1x - h2 + k. If the transverse axis is horizontal, then m= { b a . If it is vertical, then m= { a b . x y 3 22 25 0 1 (–3, –2) (–3, 1) (–3, –5) – = 1 9 (y + 2)2 4 (x + 3)2 Figure 33 EXAMPLE 3 Graphing a Hyperbola Translated Away from the Origin Graph 1y + 222 9 - 1x + 322 4 = 1. Give the equations of the asymptotes, and the domain and range. SOLUTION This equation represents a hyperbola centered at 1-3, -22. For this vertical hyperbola, a = 3 and b = 2. The x-values of the vertices are -3. Locate the y-values of the vertices by taking the y-value of the center, -2, and adding and subtracting 3. Thus, the vertices are 1-3, 12 and 1-3, -52. The asymptotes have slopes { 3 2 and pass through the center 1-3, -22. The equations of the asymptotes can be found using the point-slope form. 3y - 1-224 = { 3 2 3x - 1-324 Point-slope form: y - y1 = m1x - x12; Let y1 = -2, m= { 3 2 , and x1 = -3. y = { 3 2 1x + 32 - 2 Solve for y. The graph is shown in Figure 33. The domain of the relation is 1-∞, ∞2, and the range is 1-∞, -54 ´31, ∞2. S Now Try Exercise 21.
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