If S and S^{prime} are the foci of an ellipse | vedclass

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(C) Let the coordinates of the foci of the ellipse be $S(ae, 0)$ and $S^{\prime}(-ae, 0)$. The coordinates of the end of the minor axis $B$ are $(0, b

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

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(C) Let the coordinates of the foci of the ellipse be $S(ae, 0)$ and $S^{\prime}(-ae, 0)$. The coordinates of the end of the minor axis $B$ are $(0, b)$.Since $	riangle SBS^{\prime}$ is a right-angled triangle at $B$,we have $SB^2 + S^{\prime}B^2 = (SS^{\prime})^2$.The distance $SB = \sqrt{(ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}$.Similarly,$S^{\prime}B = \sqrt{(-ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}$.Also,$SS^{\prime} = 2ae$.Substituting these into the Pythagorean theorem:$(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$$b^2 = a^2e^2$$\frac{b^2}{a^2} = e^2$We know that for an ellipse,$b^2 = a^2(1 - e^2)$,so $\frac{b^2}{a^2} = 1 - e^2$.Equating the two expressions for $\frac{b^2}{a^2}$:$e^2 = 1 - e^2$$2e^2 = 1$$e^2 = \frac{1}{2}$$e = \frac{1}{\sqrt{2}}$

If $S$ and $S^{\prime}$ are the foci of an ellipse,$B$ is one end of the minor axis and $\angle SBS^{\prime} = 90^{\circ}$,then the eccentricity of that ellipse is

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