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

Focus (ae, 0)

$F (2 \sqrt{2}, 0)$

Line L: $y = mx + c$ pass $(1,0)$

$o = m + C$…….(1)

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

$m^{2}=4 m^{2}-4$

$4=3 m^{2}$

$\frac{2}{\sqrt{3}}= m \quad($ 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,$\left|\frac{1}{2}\right| \begin{array}{lllll}0 & x _{1} & 2 \sqrt{2} & x _{2} & 0 \\ 0 & y _{1} & 0 & y _{2} & 0\end{array} \mid$

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

$=\sqrt{2}\left| y _{2}- y _{1}\right|=\frac{(\sqrt{2}) \sqrt{12+16}}{111}$

$=\sqrt{56}$

$=2 \sqrt{14}$