Question

Consider a branch of 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 end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then 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

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

Given equation can be re-written as

$x^2-2y^2-2\sqrt{2}x-4\sqrt{2}y-6=0$

$\Rightarrow x^2-2\sqrt{2}x+{\left(\sqrt{2}\right)}^2-{\left(\sqrt{2}\right)}^2-2y^2-4\sqrt{2}y-6=0$

$\Rightarrow {(x-\sqrt{2})}^2-{\left(\sqrt{2}\right)}^2-2y^2-4\sqrt{2}y-6=0$

$\Rightarrow {(x-\sqrt{2})}^2-2-2y^2-4\sqrt{2}y-6=0$

$\Rightarrow {(x-\sqrt{2})}^2-2y^2-4\sqrt{2}y-8=0$

$\Rightarrow {(x-\sqrt{2})}^2-2(y^2+2\sqrt{2}y+4)=0$

$\Rightarrow {(x-\sqrt{2})}^2-2(y^2+2\sqrt{2}y+2-2+4)=0$

$\Rightarrow {(x-\sqrt{2})}^2-2{(y+\sqrt{2})}^2=4$

$\Rightarrow \frac{{(x-\sqrt{2})}^2}{4}-\frac{{(y+\sqrt{2})}^2}{2}=1$

For point $A(x,y)$

$e=\sqrt{1+\frac{2}{4}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$

$\Rightarrow x-\sqrt{2}=2$

$\therefore x=2+\sqrt{2}$

For point $C(x,y)$

$x-\sqrt{2}=ae=\sqrt{6}$

$\Rightarrow x=\sqrt{6}+\sqrt{2}$

Now $AC=\sqrt{6}+\sqrt{2}-2-\sqrt{2}=\sqrt{6}-2$

and $BC=\frac{b^2}{a}=\frac{2}{2}=1$

Area of $△ABC=\frac{1}{2}\times (\sqrt{6}-2)\times 1=\sqrt{\frac{3}{2}}-1$ sq.units.