The point P(-2 6, 3) lies on the hyperbola x^ 2 a^ 2 - y^ 2 b^ 2 … - Vidyadip

MCQ

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 :

A
$4 \sqrt{3}$
B
$6$
C
$6 \sqrt{3}$
✓
$3 \sqrt{6}$

✓

Answer

Correct option: D.

$3 \sqrt{6}$

d
$\mathrm{P}(-2 \sqrt{6}, \sqrt{3})$ 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 $P :$

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

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

$Q R=6 \sqrt{3}$

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