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

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

Explanation for the correct option:

Step 1. Draw the figure according to the question:

As we know, equation of hyperbola is

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

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

$=\sqrt{\frac{3}{2}}$

$\therefore$Focus, $F\left(ae,0\right)=F\left(\sqrt{6},0\right)$

Step 2. Equation of tangent at $P$ to the hyperbola is $2x-y\sqrt{6}=2$

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

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

$\therefore$Area of $△QFR=\frac{1}{2}\left(\sqrt{6}-1\right)\times \frac{2}{\sqrt{6}}\left(\sqrt{6}-1\right)$

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

Hence, Option ‘D’ is Correct.