If the area of the auxiliary circle of the el | vedclass

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(B) The area of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is given by $A_{e} = \pi ab$.The auxiliary circle of this ellipse is $x^{2

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$\frac{\sqrt{3}}{2}$

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(B) The area of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is given by $A_{e} = \pi ab$.The auxiliary circle of this ellipse is $x^{2} + y^{2} = a^{2}$,and its area is $A_{c} = \pi a^{2}$.According to the problem,$A_{c} = 2 A_{e}$.Substituting the values,we get $\pi a^{2} = 2 \pi ab$,which simplifies to $a = 2b$.The eccentricity $e$ of an ellipse is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$.Substituting $a = 2b$,we get $e = \sqrt{1 - \frac{b^{2}}{(2b)^{2}}} = \sqrt{1 - \frac{b^{2}}{4b^{2}}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

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

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