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

Given that the latus rectum of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1................\left(i\right)$

subtends an angle of $60°$ at the other focus.

As, foci are $S\left(ae,0\right) \& S^{\text{'}}\left(-ae,0\right)$

End Points of latus rectum through $S$ are given $A\left(ae, \frac{b^2}{a}\right), B\left(ae, -\frac{b^2}{a}\right)$

Given, $AB$ subtends ${60}^0$ at $S^{\text{'}}$

Angle between $S^{\text{'}}A$ with $x$-axis will be ${30}^0$.

Slope of $S^{\text{'}}A=m_1=\frac{b^2}{2a^{2}e}=\tan 30°$

$\Rightarrow \frac{b^2}{2a^{2}e}=\tan {30}^0=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{b^2}{a^2}=\frac{2e}{\sqrt{3}}........\left(ii\right)$

Also, we know that

$b^2=a^2\left(e^2-1\right)$

$\Rightarrow \frac{b^2}{a^2}=e^2-1$

Putting value of  $\frac{b^2}{a^2}$ from $\left(ii\right)$

$\Rightarrow e^2-1=\frac{2e}{\sqrt{3}}$

$\Rightarrow \sqrt{3}{e}^2-2e-\sqrt{3}=0$

$\Rightarrow$$\sqrt{3}{e}^2-3e+e-\sqrt{3}=0$

$\Rightarrow \left(e-\sqrt{3}\right)\left(\sqrt{3}e+1\right)=0$

$\therefore e=\sqrt{3},\frac{-1}{\sqrt{3}}$

But eccentricity can not be negative.

Hence, value of eccentricity $e=\sqrt{3}$