If B is the end point of the minor axis of th | vedclass

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(A) The equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$.The coordinates of the foci are $S(-ae, 0)$ and $S^{\prime}(ae, 0)$

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$\frac{1}{2}$

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(A) The equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$.The coordinates of the foci are $S(-ae, 0)$ and $S^{\prime}(ae, 0)$.The end point of the minor axis is $B(0, b)$.Since $\Delta SBS^{\prime}$ is an equilateral triangle,the angle $\angle BSO = 60^{\circ}$ in the right-angled triangle $\Delta SOB$.In $\Delta SOB$,$	an 60^{\circ} = \frac{OB}{OS} = \frac{b}{ae}$.Thus,$\sqrt{3} = \frac{b}{ae}$,which implies $b^{2} = 3a^{2}e^{2}$.Using the relation $b^{2} = a^{2}(1 - e^{2})$,we have $a^{2}(1 - e^{2}) = 3a^{2}e^{2}$.$1 - e^{2} = 3e^{2} \implies 4e^{2} = 1 \implies e^{2} = \frac{1}{4}$.Therefore,$e = \frac{1}{2}$.

If $B$ is the end point of the minor axis of the ellipse $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$ $(a > b)$ and $S$ and $S^{\prime}$ are the foci of the ellipse such that $\Delta SBS^{\prime}$ is an equilateral triangle,then the eccentricity $e$ is:

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