In an ellipse,let B be one end of the minor a | vedclass

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.The coordinates of the foci are $F(ae, 0)$ and $F'(-ae, 0

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

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a > b$.The coordinates of the foci are $F(ae, 0)$ and $F'(-ae, 0)$,and the end of the minor axis is $B(0, b)$.Given $\angle FBF' = 90^{\circ}$,the triangle $	riangle FBF'$ is a right-angled triangle at $B$.Since $O$ is the origin $(0,0)$,$OB \perp FF'$.In $	riangle OBF$,$\angle OBF = 45^{\circ}$ (since $	riangle FBF'$ is isosceles with $BF = BF'$).Thus,$	an(45^{\circ}) = \frac{OF}{OB} = \frac{ae}{b}$.$1 = \frac{ae}{b} \Rightarrow b = ae$.We know that $b^2 = a^2(1 - e^2)$.Substituting $b = ae$,we get $(ae)^2 = a^2(1 - e^2)$.$a^2e^2 = a^2 - a^2e^2$.$2a^2e^2 = a^2$.$e^2 = \frac{1}{2} \Rightarrow e = \frac{1}{\sqrt{2}}$.

In an ellipse,let $B$ be one end of the minor axis,$F$ and $F'$ be the foci,and $\angle FBF' = 90^{\circ}$. Then the eccentricity of the ellipse is:

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