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

To solve this problem, we start by determining the position of the focus \(S\) of the given hyperbola \(\frac{x^2}{3} - \frac{y^2}{5} = 1\). For a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the coordinates of the foci are \((\pm c, 0)\) where \(c = \sqrt{a^2 + b^2}\). Here, \(a^2 = 3\) and \(b^2 = 5\), so \(c = \sqrt{3 + 5} = \sqrt{8} = 2\sqrt{2}\). Since \(S\) is on the positive x-axis, \(S = (2\sqrt{2}, 0)\).

Next, we need to find the radius of circle \(C\) with center \(A(\sqrt{6}, \sqrt{5})\) that passes through \(S = (2\sqrt{2}, 0)\). The radius \(r\) is the distance from \(A\) to \(S\):

\[ r = \sqrt{(\sqrt{6} - 2\sqrt{2})^2 + (\sqrt{5} - 0)^2} = \sqrt{(\sqrt{6} - 2\sqrt{2})^2 + 5} \]

First, simplify \((\sqrt{6} - 2\sqrt{2})^2\):

\[ (\sqrt{6} - 2\sqrt{2})^2 = 6 - 4\sqrt{12} + 8 = 14 - 4\sqrt{12} \]

Therefore,

\[ r = \sqrt{14 - 4\sqrt{12} + 5} = \sqrt{19 - 4\sqrt{12}} \]

Now, we need to find point \(B\), which is diametrically opposite to \(S\) in circle \(C\):

The coordinates of \(B\) are derived using the midpoint formula, knowing \(A\) is the midpoint of \(SB\):

\( x_B = 2\sqrt{6} - 2\sqrt{2} \) and \( y_B = 2\sqrt{5} \)

Let \(B = (x_B, y_B) = (2\sqrt{6} - 2\sqrt{2}, 2\sqrt{5})\).

Finally, find the area of triangle \(OSB\) using determinant method for vertices \(O(0,0)\), \(S(2\sqrt{2}, 0)\), and \(B(2\sqrt{6} - 2\sqrt{2}, 2\sqrt{5})\):

\[\text{Area} = \frac{1}{2} \left| 0(0 - 2\sqrt{5}) + 2\sqrt{2}(2\sqrt{5} - 0) + (2\sqrt{6} - 2\sqrt{2})(0 - 0) \right| \]

\[= \frac{1}{2} \left| 4\sqrt{10} \right| = 2\sqrt{10} \]

Square of the area is:

\((2\sqrt{10})^2 = 40\)

This value falls within the specified range. Hence, the square of the area of triangle \(OSB\) is 40.

Answer 2

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.