Question

What is the eccentricity if the latus rectum of a hyberbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ focus subtends an angle of ${60}^0$ at other focus

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

Answer 1

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

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

let the eccentricity$:e$

foci: $F(ae,0)$ and $F^{\prime }(-ae,0)$

length of $L.R=\frac{2b^2}{a}$

equation of L.R:$x=ae$, Endpoints of L.R are $L(ae,\frac{b^2}{a})\&L^{\prime }(ae,\frac{-b^2}{a})$. As Major axis bisects $\angle L{F}^{\prime }{L}^{\prime }\Rightarrow \angle L{F}^{\prime }F={30}^0$ and $\angle LF{F}^{\prime }={90}^0$

$\Rightarrow \tan {30}^0=\frac{1}{2}\left(\frac{2b^2}{a}\right)\times \frac{1}{2ae}$

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

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