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 :
Correct Option: , 3
$\mathrm{P}(-2 \sqrt{6}, \sqrt{3})$ 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 :
$y-\sqrt{3}=\sqrt{2}(x+2 \sqrt{6})$
$\Rightarrow \mathrm{R}=(0,5 \sqrt{3})$
$\mathrm{QR}=6 \sqrt{3}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.