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If the latus rectum of an ellipse is equal to half of minor axis, then find its ecentricity.
Solution:
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Verified Answer
Let the equation of elipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\therefore \quad$ Length of minor axis $=2 b$
$\&$ length of latus rectum $=\frac{2 b^2}{a}$
$$
\begin{aligned}
&\therefore \frac{2 b^2}{a}=\frac{1}{2}(2 b) \\
&\Rightarrow a=2 b
\end{aligned}
$$
Eccentricity can be calculated as:
$$
\begin{aligned}
&b^2=a^2\left(1-e^2\right)=4 b^2\left(1-e^2\right) \\
&\Rightarrow e^2=1-\frac{1}{4}=\frac{3}{4} \Rightarrow e=\frac{\sqrt{3}}{2}
\end{aligned}
$$
$\therefore \quad$ Length of minor axis $=2 b$
$\&$ length of latus rectum $=\frac{2 b^2}{a}$
$$
\begin{aligned}
&\therefore \frac{2 b^2}{a}=\frac{1}{2}(2 b) \\
&\Rightarrow a=2 b
\end{aligned}
$$
Eccentricity can be calculated as:
$$
\begin{aligned}
&b^2=a^2\left(1-e^2\right)=4 b^2\left(1-e^2\right) \\
&\Rightarrow e^2=1-\frac{1}{4}=\frac{3}{4} \Rightarrow e=\frac{\sqrt{3}}{2}
\end{aligned}
$$
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