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Q.
If the area of the auxiliary 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
Equation of auxiliary circle of the ellipse,
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { is } x^{2}+y^{2}=a^{2}$
$\therefore $ Area of auxiliary circle $=\pi a^{2}$
and area of an ellipse $=\pi \,a b$
Now, according to the question
$\pi a^{2}=2(\pi \,a b) $
$\Rightarrow a=2 b $
$\Rightarrow b=\frac{a}{2}\,\,\,\,\,\,...(i)$
$\therefore $ Eccentricity of an ellipse
$=\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\sqrt{\frac{a^{2}-\frac{a^{2}}{4}}{a^{2}}}\,\,\, $ [from Eq.(i)]
$=\sqrt{\frac{3 a^{2}}{4 a^{2}}}=\frac{\sqrt{3}}{2}$