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.