Let frac{x^2}{a^2}+frac{y^2}{b^2}=1, ab be a | vedclass

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(D) Given the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$.Since $	riangle ABC$ is equilateral,the angle $\angle BAC = 60^{\circ}$.Due to t

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$\frac{1}{2}$

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(D) Given the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$.Since $	riangle ABC$ is equilateral,the angle $\angle BAC = 60^{\circ}$.Due to the symmetry of the ellipse,the line $AO$ (the $y$-axis) bisects $\angle BAC$,so $\angle OAF_2 = 30^{\circ}$.In the right-angled $	riangle AOF_2$,we have $	an(30^{\circ}) = \frac{OF_2}{OA}$.Here,$OF_2 = ae$ and $OA = b$.So,$\frac{1}{\sqrt{3}} = \frac{ae}{b}$,which implies $b = \sqrt{3}ae$.Squaring both sides,$b^2 = 3a^2e^2$.Using the relation $b^2 = a^2(1-e^2)$,we get $a^2(1-e^2) = 3a^2e^2$.Dividing by $a^2$,we have $1-e^2 = 3e^2$,which simplifies to $4e^2 = 1$.Thus,$e^2 = \frac{1}{4}$,and since $e>0$,$e = \frac{1}{2}$.

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