JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 24)
Let $$\mathrm{S}$$ be the focus of the hyperbola $$\frac{x^2}{3}-\frac{y^2}{5}=1$$, on the positive $$x$$-axis. Let $$\mathrm{C}$$ be the circle with its centre at $$\mathrm{A}(\sqrt{6}, \sqrt{5})$$ and passing through the point $$\mathrm{S}$$. If $$\mathrm{O}$$ is the origin and $$\mathrm{SAB}$$ is a diameter of $$\mathrm{C}$$, then the square of the area of the triangle OSB is equal to __________.
Answer
40
Explanation
$$\begin{aligned} & \frac{x^2}{3}-\frac{y^2}{5}=1 \\ & 5=3\left(e^2-1\right) \Rightarrow e=\sqrt{\frac{8}{3}} \\ & S \equiv(2 \sqrt{2}, 0) \end{aligned}$$
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$$A$$ is mid-point of $$B S$$
$$\Rightarrow \quad B(2 \sqrt{6}-2 \sqrt{2}, 2 \sqrt{5})$$
$$\Delta (OSB) = \left| {{1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr {2\sqrt 2 } & 0 & 1 \cr {2\sqrt 6 - 2\sqrt 2 } & {2\sqrt 5 } & 1 \cr } } \right|} \right| = 2\sqrt {10} $$<./p>
$$(\Delta(O S B))^2=40$$
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