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If the area of the auxillary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)$ is twice the area of the ellipse, then the eccentricity of the ellipse is
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Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{2}$
Given, ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, whose area is $2 \pi a b$. The auxillary circle of the ellipse is $x^{2}+y^{2}=a^{2}$ whose area is $\pi a^{2}$.
Given that, $\pi a^{2}=2 \pi a b$
$$
a=2 b
$$
Now, eccentricity of ellipse
$$
\begin{aligned}
&=\sqrt{1-\frac{b^{2}}{a^{2}}} \\
&=\sqrt{1-\frac{b^{2}}{4 b^{2}}} \\
&=\sqrt{\frac{4-1}{4}}=\frac{\sqrt{3}}{2}
\end{aligned}
$$
Given that, $\pi a^{2}=2 \pi a b$
$$
a=2 b
$$
Now, eccentricity of ellipse
$$
\begin{aligned}
&=\sqrt{1-\frac{b^{2}}{a^{2}}} \\
&=\sqrt{1-\frac{b^{2}}{4 b^{2}}} \\
&=\sqrt{\frac{4-1}{4}}=\frac{\sqrt{3}}{2}
\end{aligned}
$$
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