Let $S$ be the focus of the hyperbola $\dfrac{x^2}{3}-\dfrac{y^2}{5}=1$, on the positive $x$-axis. Let $C$ be the circle with its centre at $A(\sqrt{6},\sqrt{5})$ and passing through the point $S$. If $O$ is the origin and $SAB$ is a diameter of $C$, then the square of the area of the triangle $OSB$ is equal to
Step-by-Step Solution
Key Concept: Hyperbola: $a^2=3$, $b^2=5$, $c^2=8$. Focus $S=(\sqrt{8},0)=(2\sqrt{2},0)$. Centre of $C$ is $A(\sqrt{6},\sqrt{5})$. $SAB$ is diameter so $B=2A-S=(2\sqrt{6}-2\sqrt{2},2\sqrt{5})=(2\sqrt{8}-\sqrt{6},2\sqrt{5})$... from solution $B=(2\sqrt{8}-\sqrt{6},2\sqrt{5})$.
$S=(2\sqrt{2},0)$, $B=(2\sqrt{8}-\sqrt{6},2\sqrt{5})$. Area $=\frac{1}{2}\cdot OS\cdot h=\frac{1}{2}\sqrt{8}\cdot2\sqrt{5}=\sqrt{40}$. Square of area $=40$.
Correct Answer: 40