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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 the ellipse is $x^2 + y^2 = a

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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 the ellipse is $x^2 + y^2 = a^2$,and its area is $A_{c} = \pi a^2$.According to the problem,$A_{c} = 2 A_{e}$,so $\pi a^2 = 2 \pi ab$.Dividing by $\pi a$ (since $a 
eq 0$),we get $a = 2b$.The eccentricity $e$ of the 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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