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

Answer 1

The correct option is A

$\sqrt{\frac{3}{2}}-1$ sq unit

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

For point A ,
$\Rightarrow x-\sqrt{2}=2$
$\Rightarrow x=2+\sqrt{2}$

Co-ordinates of A = ($2+\sqrt{2},0$)

For point $C,x-\sqrt{2}=ae=\sqrt{6}\Rightarrow x=\sqrt{6}+\sqrt{2}$

Co-ordinates of A = ($\sqrt{6}+\sqrt{2},0$)
Now, $AC=\sqrt{6}+\sqrt{2}-2-\sqrt{2}=\sqrt{6}-2$
and $BC=\frac{b^2}{2}=\frac{2}{2}=1$
$\therefore$ Area of $\Delta ABC=\frac{1}{2}\times (\sqrt{6}-2)\times 1=\sqrt{\frac{3}{2}}-1squnit$