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

• A

  $\frac{6}{5}$
• B

  $\frac{50}{3}$
• C

  $\frac{10}{3}$
• Correct answer
  D

  $\frac{18}{5}$

Solution

$\begin{aligned} & 2 \mathrm{be}=8 \\ & \mathrm{be}=4\end{aligned}$

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

Hence, the correct answer is option: 'D'.