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(C) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The foci are $F_1(-ae, 0)$ and $F_2(ae, 0)$. The latus rectum through $F_1$ has endp

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$\sqrt{3}$

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(C) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The foci are $F_1(-ae, 0)$ and $F_2(ae, 0)$. The latus rectum through $F_1$ has endpoints $A(-ae, b^2/a)$ and $A'(-ae, -b^2/a)$.The angle subtended by the latus rectum at the other focus $F_2$ is $60^{\circ}$.Considering the right-angled triangle formed by the focus $F_2$,the midpoint of the latus rectum $B(-ae, 0)$,and one endpoint $A(-ae, b^2/a)$,the angle at $F_2$ is $30^{\circ}$.In $	riangle ABF_2$,we have $AB = \frac{b^2}{a}$ and $BF_2 = ae - (-ae) = 2ae$.Thus,$	an 30^{\circ} = \frac{AB}{BF_2} = \frac{b^2/a}{2ae} = \frac{b^2}{2a^2e}$.Since $\frac{1}{\sqrt{3}} = \frac{b^2}{2a^2e}$ and $b^2 = a^2(e^2 - 1)$,we substitute $b^2$:$\frac{1}{\sqrt{3}} = \frac{a^2(e^2 - 1)}{2a^2e} = \frac{e^2 - 1}{2e}$.This gives $2e = \sqrt{3}(e^2 - 1)$,or $\sqrt{3}e^2 - 2e - \sqrt{3} = 0$.Solving the quadratic equation: $e = \frac{2 \pm \sqrt{4 - 4(\sqrt{3})(-\sqrt{3})}}{2\sqrt{3}} = \frac{2 \pm \sqrt{4 + 12}}{2\sqrt{3}} = \frac{2 \pm 4}{2\sqrt{3}}$.Since $e > 1$,we take $e = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.

If the latus rectum of a hyperbola through one focus subtends an angle of $60^{\circ}$ at the other focus,then its eccentricity is

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