Let S and S^{prime} be the foci of an ellipse | vedclass

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(A) The coordinates of the foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$,and the end of the minor axis is $B(0, b)$.Since $	riangle SBS^{\prime}$ is a

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$\frac{1}{\sqrt{2}}$

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(A) The coordinates of the foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$,and the end of the minor axis is $B(0, b)$.Since $	riangle SBS^{\prime}$ is an isosceles right-angled triangle with the right angle at $B$,we have $SB = S^{\prime}B$ and $SB^2 + S^{\prime}B^2 = SS^{\prime 2}$.Given $S(ae, 0)$,$S^{\prime}(-ae, 0)$,and $B(0, b)$,the distance $SB = \sqrt{(ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}$.Similarly,$S^{\prime}B = \sqrt{(-ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}$.The distance $SS^{\prime} = 2ae$.Applying the Pythagorean theorem: $SB^2 + S^{\prime}B^2 = SS^{\prime 2}$.$(a^2e^2 + b^2) + (a^2e^2 + b^2) = (2ae)^2$.$2(a^2e^2 + b^2) = 4a^2e^2$.$a^2e^2 + b^2 = 2a^2e^2$.$b^2 = a^2e^2$.Since $b^2 = a^2(1-e^2)$,we have $a^2(1-e^2) = a^2e^2$.$1-e^2 = e^2$.$2e^2 = 1$.$e^2 = \frac{1}{2}$.$e = \frac{1}{\sqrt{2}}$.

Let $S$ and $S^{\prime}$ be the foci of an ellipse and $B$ be one end of its minor axis. If $\triangle SBS^{\prime}$ is an isosceles right-angled triangle,then the eccentricity of the ellipse is

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