Question

Find the equation of an ellipse whose axes lie along the coordinates axes, which passes through the point (- 3, 1) and has eccentricity equal to \(\sqrt{\cfrac25}\).

Answer 1

Given that we need to find the equation of the ellipse whose eccentricity is \(\sqrt{\cfrac25}\) and passes through (-3, 1).

Let us assume the equation of the ellipse is 

We know that eccentricity of the ellipse is

⇒ 5a² - 5b² = 2a²

⇒ 5b² = 3a²

⇒ a² = \(\cfrac{5b^2}3\).....(2)

Substituting the point (- 3,1) in (1) we get,

⇒ 5b² = 32

⇒ b² = \(\cfrac{32}5\)

From (2),

The equation of the ellipse is

⇒ 3x² + 5y² = 32

∴ The equation of the ellipse is 3x² + 5y² = 32.