B is an extremity of the minor axis of an ell | vedclass

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(B) 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 $S(ae, 0)$ and $S^{\prim

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

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(B) 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 $S(ae, 0)$ and $S^{\prime}(-ae, 0)$.The extremity of the minor axis is $B(0, b)$.The slope of $SB$ is $m_1 = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.The slope of $S^{\prime}B$ is $m_2 = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.Since $\angle SBS^{\prime} = 90^{\circ}$, the product of the slopes is $-1$, so $m_1 	imes m_2 = -1$.$(-\frac{b}{ae}) 	imes (\frac{b}{ae}) = -1$.$\frac{b^2}{a^2 e^2} = 1 \Rightarrow b^2 = a^2 e^2$.Using the relation $b^2 = a^2(1 - e^2)$, we get $a^2(1 - e^2) = a^2 e^2$.$1 - e^2 = e^2$ $\Rightarrow 2e^2 = 1$ $\Rightarrow e^2 = \frac{1}{2}$.Since eccentricity $e > 0$, we have $e = \frac{1}{\sqrt{2}}$.

$B$ is an extremity of the minor axis of an ellipse whose foci are $S$ and $S^{\prime}$. If $\angle SBS^{\prime}$ is a right angle, then the eccentricity of the ellipse is

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