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(D) The equation of the auxiliary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $x^{2}+y^{2}=a^{2}$.Area of the auxiliary circl

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

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(D) The equation of the auxiliary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $x^{2}+y^{2}=a^{2}$.Area of the auxiliary circle $= \pi a^{2}$.Area of the ellipse $= \pi ab$.According to the problem,the area of the auxiliary circle is twice the area of the ellipse:$\pi a^{2} = 2(\pi ab)$$a^{2} = 2ab$$a = 2b \Rightarrow b = \frac{a}{2}$.The eccentricity $e$ of the ellipse is given by:$e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$$e = \sqrt{1 - \frac{(a/2)^{2}}{a^{2}}}$$e = \sqrt{1 - \frac{a^{2}/4}{a^{2}}}$$e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

If the area of the auxiliary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (where $a > b$) is twice the area of the ellipse,then the eccentricity of the ellipse is

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