Question

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 latusrectum. If $C$ is the focus of the hyperbola nearest to the point $A$, then the area of the $\Delta ABC$ is

• A. $1-\sqrt{\frac{2}{3}}$ sq unit
• B. $\sqrt{\frac{3}{2}}-1$ sq unit
• C. $1+\sqrt{\frac{2}{3}}$ sq unit
• D. $\sqrt{\frac{3}{2}}+1$sq unit

Answer 1

Answer: (B)

Given equation can be rewritten as focal chord
$\frac{(x-\sqrt{2}{)}^2}{4}-\frac{(y+\sqrt{2}{)}^2}{2}=1$
For point $A(x,y),e=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}}$
$\Rightarrow x-\sqrt{2}=2\Rightarrow x=2+\sqrt{2}$

For point $C(x,y),x-\sqrt{2}=\alpha e=\sqrt{6}\Rightarrow x=\sqrt{6}+\sqrt{2}$
Now, $AC=\sqrt{6}+\sqrt{2}-2-\sqrt{2}=\sqrt{6}-2$
and $BC=\frac{b^2}{a}=\frac{2}{2}=1$
$\therefore$ Area of $△ABC=\frac{1}{2}\times (\sqrt{6}-2)\times 1=\sqrt{\frac{3}{2}}-1$ sq unit