Question

If the latus rectum subtends a right angle at the centre of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then eccentricity $e$ is

• A. $\frac{\sqrt{3}+1}{2}$
• B. $\frac{\sqrt{5}+1}{2}$
• C. $\sqrt{\frac{2}{3}}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

The correct option is B $\frac{\sqrt{5}+1}{2}$


The coordinates of any end of latus rectum is $(ae,\frac{b^2}{a})$
From figure, we have
$ae=\frac{b^2}{a}\Rightarrow a^{2}e=a^2(e^2-1)\Rightarrow e=e^2-1$
$\Rightarrow e=\frac{\sqrt{5}+1}{2}$

Alternate Solution:
For the given hyperbola equation of latus rectum will be
$x=\pm ae$
Homogenizing with equation of hyperbola, we get
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=\frac{x^2}{a^2{e}^2}$
$\because$ Pair of lines are perpendicular to each other.
$\therefore$ Coeff. of $x^2+$ coeff. of $y^2=0$
$\Rightarrow \frac{1}{a^2}-\frac{1}{a^2{e}^2}-\frac{1}{b^2}=0\Rightarrow \frac{1}{a^2}-\frac{1}{a^2{e}^2}-\frac{1}{a^2(e^2-1)}=0\Rightarrow e=\pm (e^2-1)\therefore e=\frac{\sqrt{5}+1}{2}(\because e>1)$