Question

$A$ is the vertex of the hyperbola $x^2-2y^2-2\sqrt{5}x-4\sqrt{2}y-3=0$, $B$ is one of the end points of latus rectum and $C$ is the focus of the hyperbola. If $A,B$ and $C$ lies on same side of conjugate axis, then the area of the triangle $ABC$ is

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

Answer 1

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

Equation can be written as
$(x^2-2\sqrt{5}x+5)-2(y^2+2\sqrt{2}y+2)=3+5+(-4)$
$\Rightarrow \frac{(x-\sqrt{5}{)}^2}{4}-\frac{(y+\sqrt{2}{)}^2}{2}=1$


Eccentricity $=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}}$

Hence area of $△ABC=\frac{1}{2}\times AC\times BC$
$=\frac{1}{2}\times (ae-a)\times \frac{b^2}{a}$
$=\frac{1}{2}\times [2\times \sqrt{\frac{3}{2}}-2]\times \frac{(\sqrt{2}{)}^2}{2}$
$=\sqrt{\frac{3}{2}}-1$ sq. units