If OB is the semi-minor axis of an ellipse,F_ | vedclass

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(A) Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the equation of the ellipse.Given that $F_1B$ and $F_2B$ are perpendicular to each other.The coordi

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$\frac{1}{2}$

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(A) Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the equation of the ellipse.Given that $F_1B$ and $F_2B$ are perpendicular to each other.The coordinates are $F_1(-ae, 0)$,$F_2(ae, 0)$,and $B(0, b)$.Slope of $F_1B = \frac{b - 0}{0 - (-ae)} = \frac{b}{ae}$.Slope of $F_2B = \frac{b - 0}{0 - ae} = -\frac{b}{ae}$.Since $F_1B \perp F_2B$,the product of their slopes is $-1$:$\left(\frac{b}{ae}ight) 	imes \left(-\frac{b}{ae}ight) = -1$$-\frac{b^2}{a^2e^2} = -1 \Rightarrow b^2 = a^2e^2$.We know that for an ellipse,$b^2 = a^2(1 - e^2)$.Substituting $b^2 = a^2e^2$ into the equation:$a^2e^2 = a^2(1 - e^2)$$e^2 = 1 - e^2$$2e^2 = 1$$e^2 = \frac{1}{2}$.

If $OB$ is the semi-minor axis of an ellipse,$F_1$ and $F_2$ are its foci and the angle between $F_1B$ and $F_2B$ is a right angle,then the square of the eccentricity of the ellipse is

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