If the foci of an ellipse are (pm sqrt{5}, 0) | vedclass

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(B) Given the foci are $(\pm \sqrt{5}, 0)$,we have $ae = \sqrt{5}$.Given eccentricity $e = \frac{\sqrt{5}}{3}$.Substituting $e$ in $ae = \sqrt{5}$,we

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$4x^2 + 9y^2 = 36$

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(B) Given the foci are $(\pm \sqrt{5}, 0)$,we have $ae = \sqrt{5}$.Given eccentricity $e = \frac{\sqrt{5}}{3}$.Substituting $e$ in $ae = \sqrt{5}$,we get $a(\frac{\sqrt{5}}{3}) = \sqrt{5}$,which implies $a = 3$,so $a^2 = 9$.Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 9(1 - \frac{5}{9}) = 9(\frac{4}{9}) = 4$.The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which becomes $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Multiplying by $36$,we get $4x^2 + 9y^2 = 36$.

If the foci of an ellipse are $(\pm \sqrt{5}, 0)$ and its eccentricity is $\frac{\sqrt{5}}{3}$,then the equation of the ellipse is

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