If the lines joining the foci of the ellipse | vedclass

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(B) Let the foci be $F_1 = (ae, 0)$ and $F_2 = (-ae, 0)$, and the extremity of the minor axis be $B = (0, b)$.The lines $BF_1$ and $BF_2$ are inclined

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$\frac{1}{2}$

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(B) Let the foci be $F_1 = (ae, 0)$ and $F_2 = (-ae, 0)$, and the extremity of the minor axis be $B = (0, b)$.The lines $BF_1$ and $BF_2$ are inclined at an angle of $60^{\circ}$ to each other, so the angle each makes with the $y$-axis is $30^{\circ}$.Thus, the slope of $BF_1$ is $	an(90^{\circ} - 30^{\circ}) = 	an 60^{\circ} = \sqrt{3}$.The slope of $BF_1$ is also given by $\frac{b - 0}{0 - ae} = -\frac{b}{ae}$.Since the angle is $60^{\circ}$ between the lines, the angle between $BF_1$ and the $y$-axis is $30^{\circ}$, so $\frac{ae}{b} = 	an 30^{\circ} = \frac{1}{\sqrt{3}}$.Therefore, $\frac{b}{ae} = \sqrt{3}$, which implies $b = ae\sqrt{3}$.Using the relation $e^2 = 1 - \frac{b^2}{a^2}$, we substitute $b^2 = 3a^2e^2$:$e^2 = 1 - \frac{3a^2e^2}{a^2} = 1 - 3e^2$.$4e^2 = 1 \Rightarrow e^2 = \frac{1}{4}$.Since $e > 0$, we have $e = \frac{1}{2}$.

If the lines joining the foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$) to an extremity of its minor axis are inclined at an angle of $60^{\circ}$ to each other, then the eccentricity of the ellipse is:

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