If the latus rectum of a hyperbola subtends a | vedclass

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(D) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The centre is $(0, 0)$.The endpoints of the latus rectum are $(ae, \frac{b^2}{a})$ a

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$\sqrt{3}+1$

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(D) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The centre is $(0, 0)$.The endpoints of the latus rectum are $(ae, \frac{b^2}{a})$ and $(ae, -\frac{b^2}{a})$.The angle subtended by the latus rectum at the centre is $120^{\circ}$.Thus,the angle made by the line joining the centre to one endpoint $(ae, \frac{b^2}{a})$ with the $x$-axis is $60^{\circ}$.Therefore,$	an(60^{\circ}) = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$.Since $\sqrt{3} = \frac{b^2}{a^2e}$ and $b^2 = a^2(e^2 - 1)$,we have $\sqrt{3} = \frac{a^2(e^2 - 1)}{a^2e} = \frac{e^2 - 1}{e}$.This gives $e^2 - \sqrt{3}e - 1 = 0$.Using the quadratic formula,$e = \frac{\sqrt{3} \pm \sqrt{3 - 4(1)(-1)}}{2} = \frac{\sqrt{3} \pm \sqrt{7}}{2}$.Since $e > 1$,we take $e = \frac{\sqrt{3} + \sqrt{7}}{2}$.

If the latus rectum of a hyperbola subtends an angle of $120^{\circ}$ at its centre,then its eccentricity is

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