Question

Find the equation of an ellipse whose eccentricity is 2/3, the latus rectum is 5 and the centre is at the origin.

Answer 1

Given, that e=2/3 and latusrectum=5
i.e, `(2b^(2))/a=5rArrb^(2)=(5a)/2`
we know that, `b^(2)=a^(2)(1-e^(2))`
`rArr (5a)/2=a^(2)(1-4/9)`
`rArr 5/2=(5a)/9rArra=9//2rArra^(2)=81/4`
`rArr b^(2)=(5xx9)/(2xx2)=45/4`
So the required equation of the ellipse is `(4x^(2))/81+(4y^(2))/45=1`