If the latus rectum of a hyperbola (x2)/(a2) - (y2)/(b2) = 1...

Given that& the latus rectum of a hyperbola $\dfrac{x2}{a2} - \dfrac{y2}{b2} = 1 . . . . . . . . . . . . . . . . \left(i\right)$

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

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

End Points of latus rectum through~$S$~are given~A \left(ae, \dfrac{b2}{a}\right),B \left(ae, - \dfrac{b2}{a}\right)$</p><p>Given,~$AB$~subtends~${60}^{0}$~at~$S^{'}$</p><p>Angle between~$S^{'}A$~with~$x$-axis will be~${30}^{0}$.</p><p>Slope of~$S^{'}A = {m}_{1} = \dfrac{b2}{2a^{2}e} =\tan 30 °$</p><p>$⇒ \dfrac{b2}{2a^{2}e} =tan{30}^{0} = \dfrac{1}{3}$</p><p>$⇒ \dfrac{b2}{a2} = \dfrac{2e}{3} . . . . . . . . \left(ii\right)$</p><p>Also, we know that</p><p>$b^{2} = a^{2} \left(e^{2} - 1\right)

$⇒ \dfrac{b2}{a2} =e^{2} - 1$

Putting value of $\dfrac{b2}{a2}$~from~ \left(ii\right)

$⇒ e^{2} - 1 =$\dfrac{2e}{3}

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

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

⇒ \left(e - \sqrt{3}\right)\left(\sqrt{3}e + 1\right) = 0$</p><p>$∴ e = $\sqrt{3}, \dfrac{- 1}{3}</p><p>But~eccentricity can not be negative.</p><p>Hence, value of eccentricity~$e = $\sqrt{3}</p>$