If the latus rectum of a hyperbola frac{x^2}{ | vedclass

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(D) The foci of the hyperbola are $S(ae, 0)$ and $S'(-ae, 0)$.The endpoints of the latus rectum passing through $S$ are $A(ae, \frac{b^2}{a})$ and $B(

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$\sqrt{3}$

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(D) The foci of the hyperbola are $S(ae, 0)$ and $S'(-ae, 0)$.The endpoints of the latus rectum passing through $S$ are $A(ae, \frac{b^2}{a})$ and $B(ae, -\frac{b^2}{a})$.The angle subtended by $AB$ at $S'$ is $60^{\circ}$.Since the triangle $	riangle AS'B$ is isosceles,the line $S'S$ bisects the angle $\angle AS'B$,so $\angle AS'S = 30^{\circ}$.The slope of $S'A$ is $	an(30^{\circ}) = \frac{1}{\sqrt{3}}$.Thus,$\frac{b^2/a}{ae - (-ae)} = \frac{b^2}{2a^2e} = \frac{1}{\sqrt{3}}$.This gives $\frac{b^2}{a^2} = \frac{2e}{\sqrt{3}}$.Using the relation $b^2 = a^2(e^2 - 1)$,we have $\frac{b^2}{a^2} = e^2 - 1$.Equating the two expressions: $e^2 - 1 = \frac{2e}{\sqrt{3}}$,which simplifies to $\sqrt{3}e^2 - 2e - \sqrt{3} = 0$.Solving the quadratic equation: $(\sqrt{3}e + 1)(e - \sqrt{3}) = 0$.Since the eccentricity $e > 1$,we get $e = \sqrt{3}$.

If the latus rectum of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends an angle of $60^{\circ}$ at the other focus,then the eccentricity of the hyperbola is

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