Consider a branch of the hyperbola  $x^2-2y^2-2\sqrt{2}x-4\sqrt{2}y-6=0$ 

with vertex at the point $A.$

let $B$ be one of the end points of its latus rectum. If $C$ is the

focus of the hyperbola near $A,$ then area of the $∆ ABC$ is:

Equation of the hyperbola is $\frac{(x-\sqrt{2}{)}^2}{4}-\frac{(y+\sqrt{2}{)}^2}{2}=1$

Which can be written as 

$\frac{X^2}{4}-\frac{Y^2}{2}=1\text{ where }x=X+\sqrt{2},y=Y-\sqrt{2}\text{. }$

Eccentricity of the hyperbola is $\sqrt{\frac{4+2}{4}}=\sqrt{\frac{3}{2}}$

Area of the $\mathrm{\Delta }ABC=\frac{1}{2}\left(AC\right)\left(BC\right)$

$\begin{array}{l}=\frac{1}{2}(ae-a)\times \frac{b^2}{a}\text{ where }{a}^2=4,b^2=2\\ =\frac{1}{2}\left[2\sqrt{\frac{3}{2}}-2\right]\times \frac{2}{2}=\sqrt{\frac{3}{2}}-1\end{array}$