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

Given, parabola $y^2=4x$ and line $L: y=mx+c$,

Now plotting the diagram we get,

Now focus of hyperbola, $H:\frac{x^2}{4}-\frac{y^2}{4}=1$ will be  $\left(ae,0\right)$ $\Rightarrow F\left(2\sqrt{2},0\right)$

Now given line $L:y=mx+c$ pass focus of parabola $\left(1,0\right)$

$\Rightarrow 0=m+c \cdots \left(1\right)$

Now also given line $L$ is tangent to Hyperbola. $\frac{x^2}{4}-\frac{y^2}{4}=1$

So by using condition of tangent to hyperbola in slope form we get, $c=\pm \sqrt{a^2 m^2-l^2}$

$\Rightarrow c=\pm \sqrt{4m^2-4}$

Now from $\left(1\right)$ we get, $-m=\pm \sqrt{4 m^2-4}$

Squaring both side we get, $m^2=4 m^2-4$

$\Rightarrow 4=3{\mathrm{m}}^2$

$\Rightarrow m=\frac{2}{\sqrt{3}}$  (as $m>0$)

So, $c=-m$$\Rightarrow c=\frac{-2}{\sqrt{3}}$

So equation of line will be $\mathrm{y}=\frac{2\mathrm{x}}{\sqrt{3}}-\frac{2}{\sqrt{3}}$

Now finding intersecting point of line and parabola $y^2=4x$ we get,

$\Rightarrow {\left(\frac{2x-2}{\sqrt{3}}\right)}^2=4x$

$\Rightarrow x^2+1-2x=3x$

$\Rightarrow x^2-5x+1=0$

$\Rightarrow x_1+x_2=5 \& x_1{x}_2=1$

${\mathrm{y}}^2=4\left(\frac{\sqrt{3}\mathrm{y}+2}{2}\right)$

$y^2=2\sqrt{3}y+4$

$\Rightarrow y^2-2\sqrt{3}y-4=0$

$\Rightarrow y_1+y_2=2\sqrt{3} \& y_1{y}_2=-4$

Area of required quadrilateral will be 

$=\left|\frac{1}{2}\left|\begin{array}{lllll}0& x_1& 2\sqrt{2}& x_2& 0\\ 0& y_1& 0& y_2& 0\end{array}\right|\right|$

$=\left|\frac{1}{2}\left[-2\sqrt{2}{y}_1+2\sqrt{2}{y}_2\right]\right|$

$=\sqrt{2}\left|y_2-y_1\right|=\frac{\left(\sqrt{2}\right)\sqrt{12+16}}{111}$

$=\sqrt{56}$​​​​​​​$=2\sqrt{14}$