The eccentricity of an ellipse is 2/3,the len | vedclass

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(B) Given eccentricity $e = 2/3$ and latus rectum $L = 2b^2/a = 5$.Using the relation $e^2 = 1 - b^2/a^2$,we have $(2/3)^2 = 1 - b^2/a^2$,which implie

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$\frac{4x^2}{81} + \frac{4y^2}{45} = 1$

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(B) Given eccentricity $e = 2/3$ and latus rectum $L = 2b^2/a = 5$.Using the relation $e^2 = 1 - b^2/a^2$,we have $(2/3)^2 = 1 - b^2/a^2$,which implies $b^2/a^2 = 1 - 4/9 = 5/9$,so $b^2 = 5a^2/9$.Substituting this into the latus rectum formula: $2(5a^2/9)/a = 5$.This simplifies to $10a/9 = 5$,giving $a = 45/10 = 9/2$.Then $b^2 = 5(9/2)^2/9 = 5(81/4)/9 = 45/4$.Thus,$a^2 = 81/4$ and $b^2 = 45/4$.The equation of the ellipse is $x^2/a^2 + y^2/b^2 = 1$,which becomes $x^2/(81/4) + y^2/(45/4) = 1$,or $\frac{4x^2}{81} + \frac{4y^2}{45} = 1$.

The eccentricity of an ellipse is $2/3$,the length of the latus rectum is $5$,and the centre is $(0, 0)$. The equation of the ellipse is:

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