1027 10.4 Summary of the Conic Sections Equation Graph Description Identification 1 y −k22 a2 − 1 x −h22 b2 =1 x a a c c b b Hyperbola (h, k) F9 F y 0 Graph has vertical transverse axis. c2 = a2 + b2 Asymptotes are y = { a b1x - h2 + k. Center is 1h, k2. y2-term has a positive coefficient. x2-term has a negative coefficient. Identifying Conic Sections To recognize the type of graph that a given conic section has, we may need to transform the equation into a more familiar form. EXAMPLE 1 Determining Types of Conic Sections Identify and sketch the graph of each relation. (a) x2 = 25 + 5y2 (b) x2 - 8x + y2 + 10y = -41 (c) 4x2 - 16x + 9y2 + 54y = -61 (d) x2 - 6x + 8y - 7 = 0 SOLUTION (a) This equation does not represent a parabola because both of its variables are squared. x2 = 25 + 5y2 Given equation x2 - 5y2 = 25 Subtract 5y2. x2 25 - y2 5 = 1 Divide by 25. The equation represents a hyperbola centered at the origin, with asymptotes y = { b a x, or y = { 25 5 x. Let a = 5 and b = 25. The x-intercepts are 1{5, 02. The graph is shown in Figure 36. (b) x2 - 8x + y2 + 10y = -41 Given equation 1x2 - 8x 2 + 1y2 + 10y 2 = -41 Rewrite in anticipation of completing the square. 1x2 - 8x + 162 + 1y2 + 10y + 252 = -41 + 16 + 25 Complete the square on both x and y. 1x - 422 + 1y + 522 = 0 Factor, and add. The resulting equation is that of a “circle” with radius 0. Its graph is the single point 14, -52. See Figure 37. If we had obtained a negative number on the right (instead of 0), the equation would have had no solution at all, and there would have been no graph. y –5 5 –3 3 x x2 = 25 + 5y2 Figure 36 Divide each term by 25. y 4 –5 x 0 (4, –5) x2– 8x + y2+ 10y = –41 Figure 37
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