Question

Let the focal chord of the parabola $P : y ^2=4 x$ along the line $L : y = mx + c , m >0$ meet the parabola at the points $M$ and $N$. Let the line $L$ be a tangent to the hyperbola $H : x ^2- y ^2=4$. If $O$ is the vertex of $P$ and $F$ is the focus of $H$ on the positive $x$-axis, then the area of the quadrilateral $OMFN$ is :

• A. $2 \sqrt{6}$
• B. $2 \sqrt{14}$
• C. $4 \sqrt{6}$
• D. $4 \sqrt{14}$

Answer 1

Answer: (B)

$ O (0,0) $
$ H : \frac{ x ^2}{4}-\frac{ y ^2}{4}=1$
Focus (ae, 0)
$F (2 \sqrt{2}, 0)$
Line $L: y=m x+c$ pass $(1,0)$
$o = m + C$
Line $L$ is tangent to Hyperbola. $\frac{ x ^2}{4}-\frac{ y ^2}{4}=1$
$ C=\pm \sqrt{a^2 m^2-\ell^2} $
$ C=\pm \sqrt{4 m^2-4}$
From (1)
$-m=\pm \sqrt{4 m^2-4}$
Squaring
$m^2=4 m^2-4$
$ 4=3 m^2 $
$\frac{2}{\sqrt{3}}=m \quad(\text { as } m>0) $
$ C=-m$
$ C=\frac{-2}{\sqrt{3}} $
$ y =\frac{2 x }{\sqrt{3}}-\frac{2}{\sqrt{3}}$
$ y ^2=4 x$
$ \Rightarrow\left(\frac{2 x-2}{\sqrt{3}}\right)^2=4 x $
$\Rightarrow x ^2+1-2 x =3 x $
$ \Rightarrow x ^2-5 x +1=0$
$ y^2=4\left(\frac{\sqrt{3} y+2}{2}\right)$
$ y^2=2 \sqrt{3} y+4 $
$ \Rightarrow y^2-2 \sqrt{3} y-4=0$
Area
$ \left|\frac{1}{2}\right| \begin{vmatrix}0 & x _1 & 2 \sqrt{2} & x _2 & 0 \\0 & y _1 & 0 & y _2 & 0\end{vmatrix}||$
$=\left|\frac{1}{2}\left[-2 \sqrt{2} y _1+2 \sqrt{2} y _2\right]\right|$
$ =\sqrt{2}\left| y _2- y _1\right|=\frac{(\sqrt{2}) \sqrt{12+16}}{111}$
$ =\sqrt{56} $