Question

Find the equation of ellipse whose eccentricity is `2/3`, latus rectum is 5 and the centre is (0, 0).

Answer 1

Equations of ellipse is `x^2/a^2 + y^2/b^2` = 1  ......(i)

Given that, e = `2/3`

And latus rectum `(2b^2)/a` = 5

⇒ `b^2 = 5/2 a`   .......(ii)

We know that b² = a² (1 – e²)

⇒ `(5a)/2 = a^2(1 - 4/9)`

⇒ `5/2 = a xx 5/9`

⇒ `a = 9/2`

⇒ `a^2 = 81/4`

And b² = `5/2 xx 9/2 = 45/4`

Hence, the required equation of ellipse is `x^2(81/4) + y^2/(45/4)` = 1

⇒ `4/81 x^2 + 4/45 y^2` = 1