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The latusrectum of the ellipse is half the minor axis. Then, its eccentricity is
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1369 Upvotes
Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{2}$
Given that, $\frac{2 b^{2}}{a}=\frac{2 b}{2}$
$$
a=2 b
$$
Now,
$$
\begin{aligned}
&b^{2}=a^{2}\left(1-e^{2}\right) \\
&b^{2}=4 b^{2}\left(1-e^{2}\right) \\
&\frac{1}{4}=1-e^{2}
\end{aligned}
$$
$\Rightarrow \quad e^{2}=1-\frac{1}{4}=\frac{3}{4}$
$\Rightarrow \quad e=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$
$$
a=2 b
$$
Now,
$$
\begin{aligned}
&b^{2}=a^{2}\left(1-e^{2}\right) \\
&b^{2}=4 b^{2}\left(1-e^{2}\right) \\
&\frac{1}{4}=1-e^{2}
\end{aligned}
$$
$\Rightarrow \quad e^{2}=1-\frac{1}{4}=\frac{3}{4}$
$\Rightarrow \quad e=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$
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