# Solve this following

Question:

Consider a hyperbola $\mathrm{H}: \mathrm{x}^{2}-2 \mathrm{y}^{2}=4 .$ Let the tangent at a point $\mathrm{P}(4, \sqrt{6})$ meet the $\mathrm{x}$-axis at $\mathrm{Q}$ and latus rectum at $\mathrm{R}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right), \mathrm{x}_{1}>0 .$ If $\mathrm{F}$ is a focus of $\mathrm{H}$ which is nearer to the point $\mathrm{P}$, then the area of $\triangle Q F R$ is equal to

1. $4 \sqrt{6}$

2. $\sqrt{6}-1$

3. $\frac{7}{\sqrt{6}}-2$

4. $4 \sqrt{6}-1$

Correct Option: , 3

Solution:

$\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$

$e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{\frac{3}{2}}$

$\therefore$ Focus $\mathrm{F}(\mathrm{ae}, 0) \Rightarrow \mathrm{F}(\sqrt{6}, 0)$

equation of tangent at $\mathrm{P}$ to the hyperbola is

$2 x-y \sqrt{6}=2$

tangent meet $x$-axis at $Q(1,0)$

\& latus rectum $x=\sqrt{6}$ at $R\left(\sqrt{6}, \frac{2}{\sqrt{6}}(\sqrt{6}-1)\right)$

$\therefore$ Area of $\Delta_{\mathrm{QFR}}=\frac{1}{2}(\sqrt{6}-1) \cdot \frac{2}{\sqrt{6}}(\sqrt{6}-1)$

$=\frac{7}{\sqrt{6}}-2$