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

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.