An ellipse has OB as semi-minor axis,F and F' | vedclass

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(C) Given $\angle F'BF = 90^\circ$,which implies $F'B \perp FB$.The coordinates of the points are $B(0, b)$,$F(ae, 0)$,and $F'(-ae, 0)$.The slope of $

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

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(C) Given $\angle F'BF = 90^\circ$,which implies $F'B \perp FB$.The coordinates of the points are $B(0, b)$,$F(ae, 0)$,and $F'(-ae, 0)$.The slope of $FB$ is $m_1 = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.The slope of $F'B$ is $m_2 = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.Since $F'B \perp FB$,the product of their slopes is $-1$:$m_1 	imes m_2 = -1$$(-\frac{b}{ae}) 	imes (\frac{b}{ae}) = -1$$\frac{b^2}{a^2e^2} = 1 \implies b^2 = a^2e^2$.We know that for an ellipse,$b^2 = a^2(1 - e^2)$.Substituting $b^2 = a^2e^2$ into this equation:$a^2e^2 = a^2(1 - e^2)$$e^2 = 1 - e^2$$2e^2 = 1$$e^2 = \frac{1}{2}$$e = \frac{1}{\sqrt{2}}$.

An ellipse has $OB$ as semi-minor axis,$F$ and $F'$ as its foci,and the angle $\angle FBF'$ is a right angle. Then the eccentricity of the ellipse is

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