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

$x^2-2y^2-2\sqrt{2}x-4\sqrt{2}y-6=0$
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 $△ABC$ 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
$\left(ae,b^2/a\right)$ or $(\sqrt{6},1)$.
$\therefore AB=\sqrt{6}-2$ and $BC=1$
$\therefore$ Area of $\Delta ABC=\frac{1}{2}AB\times BC=\frac{1}{2}(\sqrt{6}-2)\times 1$
$=\sqrt{\frac{3}{2}}-1$ sq units.