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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 $\mathrm{P}$ to the hyperbola intersect its conjugate axis at the point $\mathrm{Q}$ and $\mathrm{R}$ respectively, then $QR$ is equal to :
$4 \sqrt{3}$
$6$
$6 \sqrt{3}$
$3 \sqrt{6}$
Solution
$\mathrm{P}(-2 \sqrt{6}, \sqrt{3})$ lies on hyperbola
$\mathrm{e}=\frac{\sqrt{5}}{2} \Rightarrow \mathrm{b}^{2}=\mathrm{a}^{2}\left(\frac{5}{4}-1\right) \Rightarrow 4 \mathrm{~b}^{2}=\mathrm{a}^{2}$
$\text { Put in (i) } \Rightarrow \frac{6}{\mathrm{~b}^{2}}-\frac{3}{\mathrm{~b}^{2}}=1 \Rightarrow \mathrm{b}=\sqrt{3}$
$\Rightarrow \mathrm{a}=\sqrt{12}$
$\frac{x^{2}}{12}-\frac{y^{2}}{3}=1$ $[Image]$
Tangent at $\mathrm{P}$ :
$\frac{-x}{\sqrt{6}}-\frac{y}{\sqrt{3}}=1 \Rightarrow Q(0, \sqrt{3})$
Slope of $\mathrm{T}=-\frac{1}{\sqrt{2}}$
Normal at $P :$
$y-\sqrt{3}=\sqrt{2}(x+2 \sqrt{6})$
$\Rightarrow \quad R=(0,5 \sqrt{3})$
$Q R=6 \sqrt{3}$
