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 $R\left(x_1,y_1\right),x_1>0.$ If $F$ is a focus of $H$ which is nearer to the point $P$, then the area of $\Delta QFR$ is equal to

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

Answer 1

Answer: (C)

$\frac{x^2}{4}-\frac{y^2}{2}=1$
$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{\frac{3}{2}}$
$\therefore$ Focus $F(ae,0)\Rightarrow F(\sqrt{6},0)$
equation of tangent at $P$ to the hyperbola is $2x-y\sqrt{6}=2$
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 ${\Delta }_{QFR}=\frac{1}{2}(\sqrt{6}-1)\cdot \frac{2}{\sqrt{6}}(\sqrt{6}-1)$
$=\frac{7}{\sqrt{6}}-2$