JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(S\) be the focus of the hyperbola \(\frac{x^2}{3}-\frac{y^2}{5}=1\), on the positive \(\mathrm{x}\)-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 ...........
- A \(48\)
- B \(46\)
- C \(40\)
- D \(12\)
Answer & Solution
Correct Answer
(C) \(40\)
Step-by-step Solution
Detailed explanation
Area \(=\frac{1}{2}(\mathrm{OS}) \mathrm{h}=\frac{1}{2} \sqrt{8} 2 \sqrt{5}=\sqrt{40}\)
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