Question

Find the equation of the ellipse which passes through the point (–3, 1) and has eccentricity `sqrt(2)/5`, with x-axis as its major axis and centre at the origin.

Answer 1

Let `x^2/a^2 + y^2/b^2` = 1 be the equation of the ellipse passing through the point (–3, 1).

Therefore, we have `9/a^2 + 1/b^2` = 1.

or 9b² + a² = a²b²

or 9a²(a² – e²) + a² = a²a²(1 – e²)   ......(Using b² = a²(1 – e²)

or `a^2 = 32/3`

Again `b^2 = a^2(1 - e^2) = 32/2`

`1 - 2/5 = 32/5`

Hence, the required equation of the ellipse is `x^2/(32/3) + y^2/(32/5)` = 1

or 3x² + 5y² = 32.