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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^{

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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,the area of the auxiliary circle is twice the area of the ellipse:$\pi a^{2} = 2(\pi ab)$$a^{2} = 2ab$Since $a 
eq 0$,we have $a = 2b$.The eccentricity $e$ of the ellipse is given by $e = \sqrt{1-\frac{b^{2}}{a^{2}}}$.Substituting $a = 2b$ into the formula:$e = \sqrt{1-\frac{b^{2}}{(2b)^{2}}} = \sqrt{1-\frac{b^{2}}{4b^{2}}} = \sqrt{1-\frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

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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