Consider a branch of the hyperbola $${x^2} - 2{y^2} - 2\sqrt 2 x - 4\sqrt 2 y - 6 = 0$$       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 nearest to the point $$A,$$  then the area of the triangle $$ABC$$  is :

Solution :
The given hyperbola is
$$\eqalign{ & {x^2} - 2{y^2} - 2\sqrt 2 x - 4\sqrt 2 y - 6 = 0 \cr & \Rightarrow \left( {{x^2} - 2\sqrt 2 x + 2} \right) - 2\left( {{y^2} + 2\sqrt 2 y + 2} \right) = 6 + 2 - 4 \cr & \Rightarrow {\left( {x - \sqrt 2 } \right)^2} - 2{\left( {y + \sqrt 2 } \right)^2} = 4 \cr & \Rightarrow \frac{{{{\left( {x - \sqrt 2 } \right)}^2}}}{{{2^2}}} - \frac{{{{\left( {y + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 2 } \right)}^2}}} = 1 \cr & \therefore a = 2,\,\,\,b = \sqrt 2 \cr & \Rightarrow e = \sqrt {1 + \frac{2}{4}} = \sqrt {\frac{3}{2}} \cr} $$
Clearly $$\Delta ABC$$   is a right triangle.

$$\eqalign{ & \therefore Ar\left( {\Delta ABC} \right) = \frac{1}{2} \times AC \times BC = \frac{1}{2}\left( {ae - a} \right) \times \frac{{{b^2}}}{a} \cr & = \frac{1}{2}\left( {e - 1} \right) \times {b^2} = \frac{1}{2}\left( {\sqrt {\frac{3}{2}} - 1} \right) \times 2 = \sqrt {\frac{3}{2}} - 1 \cr} $$