Let S be the focus of the hyperbola x^23-y^25=1, on the positive x-axis.

Question

Let $S$ be the focus of the hyperbola $\frac{{x}^{2}}{3}-\frac{{y}^{2}}{5}=1$, 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 \_\_\_\_\_\_\_\_\_\_.

Answer: 40

Accepted Answer

$\begin{gathered} \frac{{x}^{2}}{3}-\frac{{y}^{2}}{5}=1 \ 5=3({e}^{2}-1)\Rightarrow e=\sqrt{\frac{8}{3}} \ S\equiv (2\sqrt{2},0) \end{gathered}$

$A$ is mid-point of $BS$

$\Rightarrow \, B(2\sqrt{6}-2\sqrt{2},2\sqrt{5})$

$\Delta(OSB)=|\frac{1}{2}\begin{vmatrix} 0 \quad  0 \quad  1 \ 2\sqrt{2} \quad  0 \quad  1 \ 2\sqrt{6}-2\sqrt{2} \quad  2\sqrt{5} \quad  1 \end{vmatrix}|=2\sqrt{10}$<./p>

$(\Delta(OSB){)}^{2}=40$

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