The length of the latus-rectum of the ellipse, whose foci are $(2,5)$ and $(2,-3)$ and eccentricity is $\frac{4}{5}$, is

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Aicharya · Accepted Answer

$\begin{aligned} & 2 \mathrm{be}=8 \ & \mathrm{be}=4\end{aligned}$ $\begin{aligned} & \mathrm{b}\left(\frac{4}{5}ight)=4 \Rightarrow \mathrm{~b}=5 \ & \because \mathrm{c}^2=\mathrm{b}^2-\mathrm{a}^2 \ & 16=25-\mathrm{a}^2 \Rightarrow \mathrm{a}=3 \ & 	ext { L.R. }=\frac{2 \mathrm{a}^2}{\mathrm{~b}}=\frac{18}{5}\end{aligned}$

The length of the latus-rectum of the ellipse, whose foci are $(2,5)$ and $(2,-3)$ and eccentricity is…

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