Question

If the latus rectum of a hyperbola$(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1)$through one focus subtends ${60}^0$ at the other focus then its eccentricity is

• A. $\frac{\sqrt{3}{b}^2}{2a^2}$
• B. $\frac{b^2}{a^2}$
• C. $\frac{5b^2}{a^2}$
• D. $\frac{6b^2}{a^2}$

Answer 1

The correct option is A $\frac{\sqrt{3}{b}^2}{2a^2}$

Given Hyperbola : $(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1)$

$L{L}^{{}^{\prime }}$ subtends angle ${60}^{\circ }$ at $F^{{}^{\prime }}$.

So, by symmetry along X axis. $\angle L{F}^{{}^{\prime }}L=\angle F{F}^{{}^{\prime }}{L}^{{}^{\prime }}$

$\Rightarrow \angle L{F}^{{}^{\prime }}L=\frac{1}{2}\left({60}^{\circ }\right)={30}^{\circ }$

where $\overline{F{F}^{{}^{\prime }}}=2ae$ & $\overline{FL}=\frac{b^2}{a}\left(\frac{1}{2}LR\right)$

As LR is perpendicular to X-axis $\angle LF{F}^{{}^{\prime }}={90}^{\circ }$

So, $\tan {30}^{\circ }=\frac{\frac{b^2}{a}}{2a^{2}e}$ $(=\frac{\overline{FL}}{\overline{F{F}^{{}^{\prime }}}})$

So, $\tan {30}^{\circ }=\frac{b^2}{2a^{2}e}$ $(\tan {30}^{\circ }=\frac{1}{\sqrt{3}})$

$\frac{b^2}{2a^{2}e}=\frac{1}{\sqrt{3}}$

$e=\frac{\sqrt{3}{b}^2}{2a^2}$