Question

If \( P(10, 2\sqrt{15}) \) lies on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) and the length of the latus rectum is 8, then the square of the area of \( \Delta PS_1S_2 \) is [where \( S_1 \) and \( S_2 \) are the foci of the hyperbola].

• A. 2700
• B. 2400
• C. 1750
• D. 3600

Hint:

In problems involving hyperbolas, the length of the latus rectum and the coordinates of points on the hyperbola are key to solving for unknowns.

Answer 1

The Correct Option is A

We are required to find the square of the area of triangle \(\triangle PS_1S_2\), where \(P(10, 2\sqrt{15})\) lies on a hyperbola and \(S_1, S_2\) are its foci.

The equation of the hyperbola is:

\(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\)

The length of the latus rectum of a hyperbola is given by \(\dfrac{2b^2}{a}\). It is given that this length equals \(8\). 

\(\dfrac{2b^2}{a} = 8 \Rightarrow b^2 = 4a \quad \text{(1)}\)

Since point \(P(10, 2\sqrt{15})\) lies on the hyperbola, substitute its coordinates into the equation:

\(\dfrac{10^2}{a^2} - \dfrac{(2\sqrt{15})^2}{b^2} = 1\)

\(\dfrac{100}{a^2} - \dfrac{60}{b^2} = 1\)

Substituting \(b^2 = 4a\) from (1):

\(\dfrac{100}{a^2} - \dfrac{60}{4a} = 1\)

\(\dfrac{100}{a^2} - \dfrac{15}{a} = 1\)

Multiplying throughout by \(a^2\):

\(100 - 15a = a^2\)

\(a^2 + 15a - 100 = 0\)

\((a - 5)(a + 20) = 0\)

Since \(a > 0\), we take \(a = 5\).

Hence, \(b^2 = 4a = 20\).

The foci of the hyperbola are given by \((\pm c, 0)\), where \(c = \sqrt{a^2 + b^2}\).

\(c = \sqrt{25 + 20} = \sqrt{45} = 3\sqrt{5}\)

Thus, \(S_1(-3\sqrt{5}, 0)\) and \(S_2(3\sqrt{5}, 0)\).

Since the foci lie on the x-axis, the base of the triangle is \(S_1S_2 = 6\sqrt{5}\), and the height is the y-coordinate of point \(P = 2\sqrt{15}\).

Area of triangle:

\(\text{Area} = \dfrac{1}{2} \times 6\sqrt{5} \times 2\sqrt{15}\)

\(= 6\sqrt{75} = 6 \times 5\sqrt{3} = 30\sqrt{3}\)

Square of the area:

\((30\sqrt{3})^2 = 2700

Therefore, the square of the area of triangle \(\triangle PS_1S_2\) is:

\(\boxed{2700}\)