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 end of its latus rectum. If $S$ is the focus of the hyperbola nearest to the point $A$ then area of $△ASB$ is

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

Answer 1

The correct option is B $\sqrt{\frac{3}{2}}-1$

Given, hyperbola may be written as $\frac{(x-\sqrt{2}{)}^2}{4}-\frac{(y+\sqrt{2}{)}^2}{2}=1$
$\Rightarrow a=2,b=\sqrt{2}$
& eccentricity, $e=\sqrt{1+1/2}=\sqrt{\frac{3}{2}}$
Required area $=\frac{1}{2}(ae-a)\times \frac{b^2}{a}=\frac{1}{2}\frac{(\sqrt{3}-\sqrt{2})\times 2}{\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2})}{\sqrt{2}}$
$\Rightarrow \Delta ASB=(\sqrt{\frac{3}{2}}-1)$ .