$\text{ Consider a branch of the hyperbola }{x}^2-2y^2-2\sqrt{2}x-4\sqrt{2}y-6=0\text{ with vertex at the point }A\text{ . Let }B\text{ be one of }$

$\text{ the endpoints of its latus rectum. If }C\text{ is the focus of the hyperbola nearest to the point A, then the area of triangle ABC is}$

$\begin{array}{l}{x}^2-2y^2-2\sqrt{2}x-4\sqrt{2}y-6=0\\ \text{ or }\left(x^2-2\sqrt{2}+2\right)-2\left(y^2+2\sqrt{2}y+2\right)=4\\ \text{ or }\frac{(x-\sqrt{2}{)}^2}{4}-\frac{(y+\sqrt{2}{)}^2}{2}=1\end{array}$

$\text{Now B is one of the end points of its latus rectum and C is the focus of the hyperbola nearest to the vertex A. }$

$\mathrm{Clearly}, \mathrm{area} \mathrm{of} \mathrm{LABC} \mathrm{does} \mathrm{not} \mathrm{change} \mathrm{if} \mathrm{we} \mathrm{consider} \mathrm{similar}$

$\text{ hyperbola with center at }(0,0)\text{ or hyperbola }\frac{x^2}{4}-\frac{y^2}{2}=1$

$\text{ Here vertex is }A(2,0)\text{ . }$

$\therefore a^2{e}^2=a^2+b^2=6$

$\text{ So, one of the foci is }B(\sqrt{6},0)\text{ , point }C\text{ is }\left(ae,b^2/a\right)\text{ or }(\sqrt{6},1)\text{ . }$

$\begin{array}{l}\therefore AB=\sqrt{6}-2\text{ and }BC=1\\ \therefore \text{ Area of }\mathrm{\Delta }ABC=\frac{1}{2}AB\times BC=\frac{1}{2}(\sqrt{6}-2)\times 1\end{array}$

                                   $=\sqrt{\frac{3}{2}}-1\text{ sq.units. }$