1004 CHAPTER 10 Analytic Geometry x y 4x2 + 9y2 = 36 + = 1 0 –3 –2 2 3 9 x2 4 y2 F(Ë5 , 0) F9(–Ë5 , 0) Figure 16 x y 4 –4 –4 –8 4 8 0 F9(0, –4Ë3) F(0, 4Ë3) 4x 2 = 64 – y2 + = 1 16 x2 64 y2 Figure 17 EXAMPLE 1 Graphing Ellipses Centered at the Origin Graph each ellipse, and find the coordinates of the foci. Give the domain and range. (a) 4x2 + 9y2 = 36 (b) 4x2 = 64 - y2 SOLUTION (a) Divide each side of 4x2 + 9y2 = 36 by 36 to write the equation in standard form. x2 9 + y2 4 = 1 Standard form of an ellipse Thus, the x-intercepts are 1{3, 02 and the y-intercepts are 10, {22. The graph of the ellipse is shown in Figure 16. Because 9 74, we find the foci of the ellipse by letting a2 = 9 and b2 = 4 in c2 = a2 - b2. c2 = 9 - 4 = 5, so c = 25 By definition, c 70. The major axis is along the x-axis. Thus, the foci have coordinates A -25, 0B and A 25, 0B. The domain of this relation is 3-3, 34, and the range is 3-2, 24. (b) Write the equation 4x2 = 64 - y2 as 4x2 + y2 = 64. Then divide each side by 64 to express it in standard form. x2 16 + y2 64 = 1 Standard form of an ellipse The x-intercepts are 1{4, 02 and the y-intercepts are 10, {82. See Figure 17. Here 64 716, so a2 = 64 and b2 = 16. Use c2 = a2 - b2. c2 = 64 - 16 = 48, so c = 248 = 423 48 = 216 # 3 = 423 The major axis is on the y-axis, which means the coordinates of the foci are A0, -423 B and A0, 423 B. The domain of the relation is 3-4, 44, and the range is 3-8, 84. S Now Try Exercises 11 and 13. The graph of an ellipse is not the graph of a function. To graph the ellipse in Example 1(a) with a graphing calculator in function mode, solve for y in 4x2 + 9y2 = 36 to obtain equations of the two functions shown in Figure 18. y = 2B1 - x2 9 and y = -2B1 - x2 9 7 Each term was divided by 64. Their union is the graph of 4x 2 + 9y 2 = 36. Each term was divided by 36. −6 −4 4 6 y1 = 2 Ä 1 − x2 9 y2 = −2 Ä 1 − x2 9 Figure 18
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