The length of the latus rectum of an ellipse | vedclass

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(B) Given,the length of the latus rectum is $\frac{2b^2}{a} = \frac{18}{5}$ $\Rightarrow \frac{b^2}{a} = \frac{9}{5}$ $\Rightarrow b^2 = \frac{9}{5}a

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$\frac{x^2}{25}+\frac{y^2}{9}=1$

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(B) Given,the length of the latus rectum is $\frac{2b^2}{a} = \frac{18}{5}$ $\Rightarrow \frac{b^2}{a} = \frac{9}{5}$ $\Rightarrow b^2 = \frac{9}{5}a \dots (i)$.Given,eccentricity $e = \frac{4}{5}$.We know that $e^2 = 1 - \frac{b^2}{a^2}$,so $\frac{16}{25} = 1 - \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = 1 - \frac{16}{25} = \frac{9}{25}$.Substituting $b^2 = \frac{9}{5}a$ into the equation: $\frac{\frac{9}{5}a}{a^2} = \frac{9}{25}$ $\Rightarrow \frac{9}{5a} = \frac{9}{25}$ $\Rightarrow a = 5$.Now,$b^2 = \frac{9}{5}(5) = 9$.Thus,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

The length of the latus rectum of an ellipse is $\frac{18}{5}$ and eccentricity is $\frac{4}{5}$,then the equation of the ellipse is...

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