Let P and Q be the foci of an ellipse and let | 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 foci are $P = (-ae, 0)$ and $Q = (ae, 0)$.The end of

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$\frac{1}{2}$

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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 foci are $P = (-ae, 0)$ and $Q = (ae, 0)$.The end of the minor axis is $R = (0, b)$.Since $	riangle PQR$ is an equilateral triangle,the distance $PQ = PR = QR$.The distance $PQ = 2ae$.The distance $PR = \sqrt{(ae - 0)^2 + (0 - b)^2} = \sqrt{a^2e^2 + b^2}$.Equating $PQ^2 = PR^2$,we get $(2ae)^2 = a^2e^2 + b^2$.$4a^2e^2 = a^2e^2 + b^2 \implies 3a^2e^2 = b^2$.Using the relation $b^2 = a^2(1 - e^2)$,we have $3a^2e^2 = a^2(1 - e^2)$.$3e^2 = 1 - e^2 \implies 4e^2 = 1 \implies e^2 = \frac{1}{4}$.Thus,$e = \frac{1}{2}$.

Let $P$ and $Q$ be the foci of an ellipse and let $R$ be one end of its minor axis. If $\triangle PQR$ is an equilateral triangle,then the eccentricity of the ellipse is equal to

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