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.