Question

Let the focal chord of the parabola $P:y^2=4x$ 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 $L:y=mx+c$ pass $(1,0)$
$o=m+C$
Line $L$ is tangent to Hyperbola. $\frac{x^2}{4}-\frac{y^2}{4}=1$
$C=\pm \sqrt{a^2{m}^2-{\ell }^2}$
$C=\pm \sqrt{4m^2-4}$
From (1)
$-m=\pm \sqrt{4m^2-4}$
Squaring
$m^2=4m^2-4$
$4=3m^2$
$\frac{2}{\sqrt{3}}=m(\text{ as }m>0)$
$C=-m$
$C=\frac{-2}{\sqrt{3}}$
$y=\frac{2x}{\sqrt{3}}-\frac{2}{\sqrt{3}}$
$y^2=4x$
$\Rightarrow {\left(\frac{2x-2}{\sqrt{3}}\right)}^2=4x$
$\Rightarrow x^2+1-2x=3x$
$\Rightarrow x^2-5x+1=0$
$y^2=4\left(\frac{\sqrt{3}y+2}{2}\right)$
$y^2=2\sqrt{3}y+4$
$\Rightarrow y^2-2\sqrt{3}y-4=0$
Area
$\left|\frac{1}{2}\right|\left|\begin{array}{ccccc}0& x_1& 2\sqrt{2}& x_2& 0\\ 0& y_1& 0& y_2& 0\end{array}\right|||$
$=\left|\frac{1}{2}\left[-2\sqrt{2}{y}_1+2\sqrt{2}{y}_2\right]\right|$
$=\sqrt{2}\left|y_2-y_1\right|=\frac{(\sqrt{2})\sqrt{12+16}}{111}$
$=\sqrt{56}$