Let the focal chord of the parabola $\mathrm{P}:{\mathrm{y}}^2=4\mathrm{x}$ along the line $\mathrm{L}:\mathrm{y}=\mathrm{mx}+\mathrm{c},\mathrm{m}>0$meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola $\mathrm{H}:{\mathrm{x}}^2-{\mathrm{y}}^2=4$. 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 :

$\mathrm{P}:{\mathrm{y}}^2=4\mathrm{x}\Rightarrow \mathrm{a}=1$
$\therefore \text{ Focus of }\mathrm{P}\equiv (1,0)$
Focal chord $\mathrm{L}:\mathrm{y}=\mathrm{mx}+\mathrm{c}\text{ passes through }(1,0)$
$\therefore 0=\mathrm{m}+\mathrm{c}\Rightarrow \mathrm{c}=-\mathrm{m}$
$\mathrm{H}:{\mathrm{x}}^2-{\mathrm{y}}^2=4$,  which is a rectangular hyperbola with eccentricity $\sqrt{2}$.
$\therefore \text{ Focus of }\mathrm{H}\equiv (\mathrm{ae},0)=\mathrm{F}(2\sqrt{2},0)$
Line L is a tangent to hyperbola.

$\begin{array}{l}\therefore \mathrm{c}=\pm \sqrt{4{\mathrm{m}}^2-4}\Rightarrow -\mathrm{m}=\pm \sqrt{4{\mathrm{m}}^2-4}\\ \Rightarrow {\mathrm{m}}^2=4{\mathrm{m}}^2-4\Rightarrow 3{\mathrm{m}}^2=4\Rightarrow \mathrm{m}=\frac{2}{\sqrt{3}}(\text{ as }\mathrm{m}>0)\\ \mathrm{c}=-\mathrm{m}=\frac{-2}{\sqrt{3}}\therefore \mathrm{L}:\mathrm{y}=\frac{2}{\sqrt{3}}\mathrm{x}-\frac{2}{\sqrt{3}}\\ \mathrm{Let} \mathrm{M}\left({\mathrm{x}}_1, {\mathrm{y}}_1\right) \mathrm{and} \mathrm{N}\left({\mathrm{x}}_2, {\mathrm{y}}_2\right) \mathrm{be} \mathrm{the} \mathrm{points} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{Parabola} \mathrm{and} \mathrm{focal} \mathrm{chord}.\\ \mathrm{P}:{\mathrm{y}}^2=4\mathrm{x}\Rightarrow {\left(\frac{2\mathrm{x}-2}{\sqrt{3}}\right)}^2=4\mathrm{x}\Rightarrow \frac{4\left({\mathrm{x}}^2+1-2\mathrm{x}\right)}{3}=4\mathrm{x}\\ \Rightarrow {\mathrm{x}}^2-5\mathrm{x}+1=0\end{array}$
$\mathrm{x}=\frac{\sqrt{3}}{2}\left(\mathrm{y}+\frac{2}{\sqrt{3}}\right)$
Put $\mathrm{x}=\frac{\sqrt{3}}{2}\left(\mathrm{y}+\frac{2}{\sqrt{3}}\right)\text{ in }{\mathrm{y}}^2=4\mathrm{x}$
$\begin{array}{l}\Rightarrow {\mathrm{y}}^2=\frac{4\sqrt{3}}{2}\left(\mathrm{y}+\frac{2}{\sqrt{3}}\right)\Rightarrow {\mathrm{y}}^2-2\sqrt{3}\mathrm{y}-4=0\\ \Rightarrow {\mathrm{y}}_1+{\mathrm{y}}_2=2\sqrt{3}\Rightarrow \left|{\mathrm{y}}_1-{\mathrm{y}}_2\right|=\sqrt{28}\end{array}$
Area of OMFN is :
$\begin{array}{l}=\frac{1}{2}\left|\begin{array}{ccccc}0& {\mathrm{x}}_1& 2\sqrt{2}& {\mathrm{x}}_2& 0\\ 0& {\mathrm{y}}_1& 0& {\mathrm{y}}_2& 0\end{array}\right|=\sqrt{2}\left|{\mathrm{y}}_2-{\mathrm{y}}_1\right|\\ =\sqrt{2}\times \sqrt{28}=\sqrt{56}=2\sqrt{14}\end{array}$