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).

Answer 1

Given that we need to find the equation of the ellipse whose eccentricity is 3/4, centre at the origin and passes through (6,4).

Let us assume the equation of the ellipse is

since centre is at origin and foci on y - axis

We know that eccentricity of the ellipse is

⇒ 16b² - 16a² = 9b²

⇒ 7b² = 16a²

⇒ a² = \(\cfrac{7b^2}{16}\).....(2)

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

From (2),

⇒ a² = 43

The equation of the ellipse is

⇒ 16x² + 7y² = 688

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