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Find the equation of an ellipse whose ec...

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

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The correct Answer is:
`(4x^(2))/(45)+(4y^(2))/(81)=1 or (4x^(2))/(81)+(4y^(2))/(45)=1`
Let equation of the ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
Given that `,=(2)/(3)` and latus reactum =5
If `agtb`, then
`(2b^(2))/(a)=5rArrrb^(2)=(5a)/(2)`
Now, `b^(2)=a^(2)(1-e^(2))`
`rArr (5a)/(2)=a^(2)(1-(4)/(9))`
`rArr (5)/(2)=(5a)/(9)`
`:. a=(9)/(2)`
`:. b^(2)(5xx9)/(2xx2)=(45)/(4)`
So, the required equation of the ellipe is `(4x^(2))/(81)+(4y^(2))/(45)=1`
If `altb`, then equition of ellipse will be `(4x^(2))/(45)+(4y^(2))/(81)=1`
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