If OT is the semi-minor axis of an ellipse,A | vedclass

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

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

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(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The coordinates of the foci are $A(-ae, 0)$ and $B(ae, 0)$,and the endpoint of the semi-minor axis is $T(0, b)$. The slope of $AT$ is $m_1 = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$. The slope of $BT$ is $m_2 = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$. Since $\angle ATB = 90^{\circ}$,the product of the slopes is $-1$: $\left(\frac{b}{ae}\right) \left(-\frac{b}{ae}\right) = -1$. $\Rightarrow \frac{b^2}{a^2 e^2} = 1$ $\Rightarrow b^2 = a^2 e^2$. We know that for an ellipse,$b^2 = a^2(1 - e^2)$. Substituting $b^2 = a^2 e^2$ into the equation: $a^2 e^2 = a^2(1 - e^2)$. $e^2 = 1 - e^2 $ $\Rightarrow 2e^2 = 1 $ $\Rightarrow e^2 = \frac{1}{2} $ $\Rightarrow e = \frac{1}{\sqrt{2}}$.

If $OT$ is the semi-minor axis of an ellipse,$A$ and $B$ are its foci and $\angle ATB$ is a right angle,then the eccentricity of that ellipse is

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