Let S be the focus of the hyperbola , on the positive x-axis. Let C be the circle with its centre at 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 a diameter of C, then

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Solution

Q.
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 __________. [2024]

Ans.
(40)

$\frac{x^{3}}{3} - \frac{y^{2}}{5} = 1 \Rightarrow a = \sqrt{3}, b = \sqrt{5}$

$∴ e = \sqrt{1 + \frac{5}{3}} \Rightarrow e = \sqrt{\frac{8}{3}}$

$∴ a e = \sqrt{8} = 2 \sqrt{2} ∴ S = (2 \sqrt{2}, 0)$

A is mid-point of BS $\Rightarrow B (2 \sqrt{6} - 2 \sqrt{2}, 2 \sqrt{5)}$

Area of $△ (O S B)$ , $A =$ $|$ $\frac{1}{2}$ $| \begin{matrix} 0 & 0 & 1 \\ 2 \sqrt{2} & 0 & 1 \\ 2 \sqrt{6} - 2 \sqrt{2} & 2 \sqrt{5} & 1 \end{matrix} |$ $|$ $= 2 \sqrt{10}$

$\Rightarrow A^{2} = 40$ .

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