JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Hyperbola

Rewrite the hyperbola in standard form: $\frac{(x-\sqrt{2}{)}^2}{4}-\frac{(y+\sqrt{2}{)}^2}{2}=1$. Here $a^2=4$, $b^2=2$, so $a=2$, $b=\sqrt{2}$, and $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{\frac{3}{2}}$. The vertex $A$ is at distance $a$ from the center, and the focus $C$ nearest to $A$ is at distance $ae$ from the center. The distance $AC=ae-a=a(e-1)=2(\sqrt{\frac{3}{2}}-1)$. The semi-latus rectum is $\frac{b^2}{a}=\frac{2}{2}=1$. The area of triangle $ABC$ is $\frac{1}{2}\times AC\times \frac{b^2}{a}=\frac{1}{2}\times 2(\sqrt{\frac{3}{2}}-1)\times 1=\sqrt{\frac{3}{2}}-1$.