If theta is the angle subtended by a latus re | vedclass

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(C) The eccentricity is given by $e = \frac{2}{\sqrt{7}-\sqrt{3}}$. Rationalizing the denominator,we get $e = \frac{2(\sqrt{7}+\sqrt{3})}{7-3} = \frac

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$\frac{\sqrt{3}}{2}$

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(C) The eccentricity is given by $e = \frac{2}{\sqrt{7}-\sqrt{3}}$. Rationalizing the denominator,we get $e = \frac{2(\sqrt{7}+\sqrt{3})}{7-3} = \frac{2(\sqrt{7}+\sqrt{3})}{4} = \frac{\sqrt{7}+\sqrt{3}}{2}$.For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the coordinates of the ends of the latus rectum are $(ae, \pm \frac{b^2}{a})$.The angle $	heta$ subtended by the latus rectum at the centre $(0,0)$ is given by $	an(\frac{	heta}{2}) = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$.Using $b^2 = a^2(e^2-1)$,we have $	an(\frac{	heta}{2}) = \frac{a^2(e^2-1)}{a^2e} = \frac{e^2-1}{e}$.Given $e = \frac{\sqrt{7}+\sqrt{3}}{2}$,then $e^2 = \frac{7+3+2\sqrt{21}}{4} = \frac{10+2\sqrt{21}}{4} = \frac{5+\sqrt{21}}{2}$.So,$	an(\frac{	heta}{2}) = \frac{\frac{5+\sqrt{21}}{2} - 1}{\frac{\sqrt{7}+\sqrt{3}}{2}} = \frac{3+\sqrt{21}}{\sqrt{7}+\sqrt{3}} = \frac{\sqrt{3}(\sqrt{3}+\sqrt{7})}{\sqrt{7}+\sqrt{3}} = \sqrt{3}$.Thus,$\frac{	heta}{2} = 60^{\circ}$,which means $	heta = 120^{\circ}$.Therefore,$\sin 	heta = \sin 120^{\circ} = \frac{\sqrt{3}}{2}$.

If $\theta$ is the angle subtended by a latus rectum at the centre of the hyperbola having eccentricity $e = \frac{2}{\sqrt{7}-\sqrt{3}}$,then $\sin \theta = $

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