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 \(A\left(\sqrt{6}, \sqrt{5}\right)\) 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 1

Correct Answer: 40

Consider the hyperbola:
\[\frac{x^2}{3} - \frac{y^2}{5} = 1.\]
The focus \(S\) is located at \((\sqrt{8}, 0)\) on the positive x-axis.
The circle \(C\) has its center at \(A(\sqrt{6}, \sqrt{5})\) and passes through the point \(S\). The radius of the circle is given by:
\[r = \text{Distance between } A \text{ and } S = \sqrt{(\sqrt{6} - \sqrt{8})^2 + (\sqrt{5} - 0)^2}.\]
Simplifying:
\[r = \sqrt{(\sqrt{6} - \sqrt{8})^2 + (\sqrt{5})^2} = \sqrt{(\sqrt{6} - \sqrt{8})^2 + 5}.\]
Since \(O\) is the origin, and \(SAB\) is a diameter of circle \(C\), we can find the coordinates of point \(B\) as \((2\sqrt{8} - \sqrt{6}, 2\sqrt{5})\).
The area of triangle \(OSB\) is given by:
\[\text{Area} = \frac{1}{2} \times OS \times \text{height}.\]
Using the coordinates of \(O\), \(S\), and \(B\), we calculate:
\[\text{Area} = \frac{1}{2} \times OS \times \text{height} = \frac{1}{2} \times \sqrt{8} \times 2\sqrt{5} = \sqrt{40}.\]
The square of the area is:
\[(\sqrt{40})^2 = 40.\]
Answer: 40.