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If the eccentricity of an ellipse is $\frac{5}{8}$ and the distance between its foci is 10 , then find latus rectum of the ellipse.
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Verified Answer
Given $e=\frac{5}{8} \& 2 a e=10$
$$
\Rightarrow 2 a\left(\frac{5}{8}\right)=10 \Rightarrow a=8
$$
Also $b^2=a^2\left(1-e^2\right)$
$$
\Rightarrow b^2=64\left(1-\frac{25}{64}\right)=64-25=39
$$
$\therefore$ Latus rectum $=2 \frac{b^2}{a}=2\left(\frac{39}{8}\right)=\frac{39}{4}$
$$
\Rightarrow 2 a\left(\frac{5}{8}\right)=10 \Rightarrow a=8
$$
Also $b^2=a^2\left(1-e^2\right)$
$$
\Rightarrow b^2=64\left(1-\frac{25}{64}\right)=64-25=39
$$
$\therefore$ Latus rectum $=2 \frac{b^2}{a}=2\left(\frac{39}{8}\right)=\frac{39}{4}$
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