Question

Consider a hyperbola $H:x^2-2y^2=4.$ Let the tangent at a point $P(4,\sqrt{6})$ meet the $x-$axis at $Q$ and latus rectum at $R(x_1,y_1),x_1>0.$ If $F$ is a focus of $H$ which is nearer to the point $P,$ then the area of $\Delta QFR$ is equal to:

• A. $\sqrt{6}-1$
• B. $4\sqrt{6}-1$
• C. $4\sqrt{6}$
• D. $\frac{7}{\sqrt{6}}-2$

Answer 1

The correct option is D $\frac{7}{\sqrt{6}}-2$

Given : $H:x^2-2y^2=4$
$\Rightarrow \frac{x^2}{4}-\frac{y^2}{2}=1$


Tangent at $P(4,\sqrt{6})$
$T=0\Rightarrow 4x-\frac{\sqrt{6}y}{2}=1\Rightarrow 2x-\sqrt{6}\left(y\right)=2\cdots \left(1\right)$
Putting $y=0$, we get
$Q=(1,0)$
Eccentricity
$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}F=(\sqrt{6},0)$
Equation of latus rectum:
$x=ae=2\sqrt{\frac{3}{2}}=\sqrt{6}\cdots \left(2\right)$
Solving equations $\left(1\right)$ and $\left(2\right)$, we get
$R=(\sqrt{6},\frac{2\sqrt{6}-2}{\sqrt{6}})$

Therefore, area of $△QFR$
$\Delta =\frac{1}{2}\times QF\times FR\Rightarrow \Delta =\frac{1}{2}\times (\sqrt{6}-1)\times \left(\frac{2\sqrt{6}-2}{\sqrt{6}}\right)\Rightarrow \Delta =\frac{7}{\sqrt{6}}-2$