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 .
Given that we need to find the equation of the ellipse whose eccentricity is and passes through (- 3,1).
Let us assume the equation of the ellipse is - - - - (1) (a2>b2).
We know that eccentricity of the ellipse is
⇒
⇒
⇒ 5a2 - 5b2 = 2a2
⇒ 5b2 = 3a2
⇒ ..... - - - (2)
Substituting the point (- 3,1) in (1) we get,
⇒
⇒
⇒
⇒
⇒ 5b2 = 32
⇒
From (2),
⇒
⇒
The equation of the ellipse is
⇒
⇒
⇒
⇒ 3x2 + 5y2 = 32
∴ The equation of the ellipse is 3x2 + 5y2 = 32.