Question

If the latus rectum of a hyperbola through one focus subtends ${60}^{\circ }$ angle at the other focus, then its eccentricity e is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let $LS{L}^{{}^{\prime }}$ be a latus rectum through the focus S (ae, 0)of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. It subtends angle ${60}^{\circ }$ at
the other focus $S^{{}^{\prime }}(-ae,\theta )$

We have, $\angle L{S}^{{}^{\prime }}{L}^{{}^{\prime }}={60}^{\circ }$
$\therefore$ $\angle L{S}^{{}^{\prime }}S={30}^{\circ }$
In $\Delta L{S}^{{}^{\prime }}S$ we have
$tan{30}^{\circ }=\frac{LS}{S^{{}^{\prime }}S}\Rightarrow \frac{1}{\sqrt{3}}=\frac{b^2/a}{2ae}$
$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{b^2}{2a^{2}e}$
$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{e^2-1}{2e}$
$\Rightarrow$ $\sqrt{3}{e}^2-2e-\sqrt{3}=0$
$\Rightarrow$ $(e-\sqrt{3})(\sqrt{2}e+1)=0$
$\Rightarrow$ $\sqrt{3}$