1009 10.2 Ellipses To find the eccentricity e, use e = c a . e = 27 4 ≈0.66 Substitute for c and a. (b) Start by dividing each term of the given equation by 50. 5x2 + 10y2 = 50 Given equation 5x2 50 + 10y2 50 = 50 50 Divide by 50. x2 10 + y2 5 = 1 Write in lowest terms. For this ellipse, a2 = 10 and thus a = 210. Find c as in part (a). c2 = a2 - b2 Relationship for ellipses c2 = 10 - 5 Let a2 = 10 and b2 = 5. c = 25 Subtract. Take the positive square root. Now find e. e = 25 210 ≈0.71 e = c a S Now Try Exercises 49 and 51. Applications of Ellipses EXAMPLE 7 Applying the Equation of an Ellipse to the Orbit of a Planet The orbit of the planet Mars is an ellipse with the sun at one focus. The eccentricity of the ellipse is 0.0935, and the closest distance that Mars comes to the sun is 128.5 million mi. (Data from The World Almanac and Book of Facts.) Find the maximum distance of Mars from the sun. SOLUTION Figure 25 shows the orbit of Mars with the origin at the center of the ellipse and the sun at one focus. Mars is closest to the sun when Mars is at the right endpoint of the major axis and farthest from the sun when Mars is at the left endpoint. Therefore, the least distance is a - c, and the greatest distance is a + c. Because a - c = 128.5, it follows that c = a - 128.5. e = c a Eccentricity formula for an ellipse 0.0935 = a - 128.5 a Let e = 0.0935 as given, and c = a - 128.5. 0.0935a = a - 128.5 Multiply by a. a ≈141.8 Solve for a. Then c = 141.8 - 128.5 = 13.3 and a + c = 141.8 + 13.3 = 155.1. The maximum distance of Mars from the sun is about 155.1 million mi. S Now Try Exercise 57. x y a – c a + c c a Sun Mars NOT TO SCALE Figure 25
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