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 :

Given ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$  whose area is  $pab.$  The auxiliary circle to the given ellipse is   $x^2+y^2=a^2$ whose area is  $\pi a^2$
Given that,  $\pi a^2=2\pi ab\Rightarrow a=2b$
Now, eccentricity of ellipse,  $=\sqrt{1-\frac{b^2}{a^2}}$