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Question

Find the equation of an ellipse with its foci on y-axis, eccentricity 3/4, centre at the origin and passing through (6, 4).

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Solution

Let the equation of the ellipse bex2a2+y2b2=1 and let e=34. Also, let the foci be on the y-axis.a2=b21-e2a2=b21-916a2=716b2x2a2+7y216a2=1 ...(1)It passes through 6,4.Then 36a2+11216a2=1576+11216a2=116a2=688a2=43Now, b2=16a27b2=6887Substituting the values of a2 and b2 in eq. (1), we get:x243+7y2688=1This is the required equation of the ellipse.

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