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

Given that we need to find the equation of the ellipse whose eccentricity is , centre at the origin and passes through (6,4).



Let us assume the equation of the ellipse is - - - - (1) (a2<b2), since centre is at origin and foci on y - axis.


We know that eccentricity of the ellipse is




16b2 - 16a2 = 9b2


7b2 = 16a2


..... - - - (2)


Substituting the point (6,4) in (1) we get,






7b2 = 688



From (2),




a2 = 43


The equation of the ellipse is





16x2 + 7y2 = 688


The equation of the ellipse is 16x2 + 7y2 = 688.


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