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