# Solve this

Question:

The point $\mathrm{P}(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having eccentricity $\frac{\sqrt{5}}{2}$. If the tangent

and normal at $P$ to the hyperbola intersect its conjugate axis at the point $Q$ and $R$ respectively, then $Q R$ is equal to :

1. $4 \sqrt{3}$

2. 6

3. $6 \sqrt{3}$

4. $3 \sqrt{6}$

Correct Option: , 3

Solution:

$\mathrm{P}(-2 \sqrt{6}, \sqrt{3})$ lies on hyperbola

$\Rightarrow \frac{24}{a^{2}}-\frac{3}{b^{2}}=1$      .......(1)

$\mathrm{e}=\frac{\sqrt{5}}{2} \Rightarrow \mathrm{b}^{2}=\mathrm{a}^{2}\left(\frac{5}{4}-1\right) \Rightarrow 4 \mathrm{~b}^{2}=\mathrm{a}^{2}$

Put in (i) $\Rightarrow \frac{6}{b^{2}}-\frac{3}{b^{2}}=1 \Rightarrow b=\sqrt{3}$

$\Rightarrow a=\sqrt{12}$

Tangent at P :

$\frac{-x}{\sqrt{6}}-\frac{y}{\sqrt{3}}=1 \Rightarrow Q(0, \sqrt{3})$

Slope of $\mathrm{T}=-\frac{1}{\sqrt{2}}$

Normal at P :

$y-\sqrt{3}=\sqrt{2}(x+2 \sqrt{6})$

$\Rightarrow \mathrm{R}=(0,5 \sqrt{3})$

$\mathrm{QR}=6 \sqrt{3}$