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Find the equation of an ellipse whose vertices are `(0,+-10)`and eccentricity `e=4/5` |
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Answer» Since the vertices of the ellipse lie on the y-axis, it is a vertical ellipse. Let the required equation be `x^(2)/b^(2) + y^(2)/a^(2) = 1," where " a^(2) gt b^(2).` Its vertices are `(0, pm a) ` and therefore, a = 10. Let ` c^(2) = (a^(2) - b^(2)).` Then , ` e = c/a rArr x = ae = (10 xx 4/5) = 8`. Now, `c^(2) = (a^(2)-b^(2)) hArr b^(2) = (a^(2)-c^(2)) = (100 - 64) = 36.` `:. a^(2) = (10)^(2) = 100 and b^(2) = 36.` Hence, the required equation is ` x^(2) / 36 + y^(2)/ 100 = 1. ` |
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