Question

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 endpoints of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$, then the area of triangle $A B C$ is

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

Answer 1

Answer: (B)

$x^{2}-2 y^{2}-2 \sqrt{2} x-4 \sqrt{2} y-6=0$
or $\left(x^{2}-2 \sqrt{2}+2\right)-2\left(y^{2}+2 \sqrt{2} y+2\right)=4$
or $ \frac{(x-\sqrt{2})^{2}}{4}-\frac{(y+\sqrt{2})^{2}}{2}=1$
Now $B$ is one of the end points of its latus rectum and $C$ is the focus of the hyperbola nearest to the vertex $A$.
Clearly, area of $\triangle A B C$ does not change if we consider similar hyperbola with center at $(0,0)$
or hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$

Here vertex is $A(2, 0)$.
$\therefore a^{2} e^{2}=a^{2}+b^{2}=6$
So, one of the foci is $B(\sqrt{6}, 0)$, point $C$ is
$\left(a e, b^{2} / a\right)$ or $(\sqrt{6}, 1)$.
$\therefore A B=\sqrt{6}-2$ and $B C=1$
$\therefore $ Area of $ \Delta A B C =\frac{1}{2} A B \times B C=\frac{1}{2}(\sqrt{6}-2) \times 1$
$=\sqrt{\frac{3}{2}}-1 $ sq units.