Question

Let \( P(10, 2\sqrt{15}) \) be a point on the hyperbola \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \] whose foci are \( S \) and \( S' \). If the length of its latus rectum is \(8\), then the square of the area of \( \triangle PSS' \) is equal to

• A. 4200
• B. 900
• C. 1462
• D. 2700

Hint:

For hyperbolas, remember: Latus rectum length \( = \dfrac{2b^2}{a} \) and \( c^2 = a^2 + b^2 \).

Answer 1

The Correct Option is D

For the hyperbola \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \] the length of the latus rectum is \[ \frac{2b^2}{a} = 8 \Rightarrow b^2 = 4a. \]
Step 1: Use the given point \( P(10, 2\sqrt{15}) \).
Substitute in the equation of hyperbola: \[ \frac{100}{a^2} - \frac{60}{b^2} = 1. \] Using \( b^2 = 4a \), \[ \frac{100}{a^2} - \frac{60}{4a} = 1 \Rightarrow \frac{100}{a^2} - \frac{15}{a} = 1. \] Multiplying by \( a^2 \), \[ 100 - 15a = a^2 \Rightarrow a^2 + 15a - 100 = 0. \] \[ a = 5,\quad b^2 = 20. \]
Step 2: Find the focal distance.
\[ c^2 = a^2 + b^2 = 25 + 20 = 45 \Rightarrow c = 3\sqrt{5}. \] Distance between foci: \[ SS' = 2c = 6\sqrt{5}. \]
Step 3: Find area of \( \triangle PSS' \).
The base \( SS' \) lies on the \(x\)-axis and height of point \(P\) is \(2\sqrt{15}\). \[ \text{Area} = \frac{1}{2} \times 6\sqrt{5} \times 2\sqrt{15} = 6\sqrt{75} = 30\sqrt{3}. \] \[ (\text{Area})^2 = (30\sqrt{3})^2 = 2700. \]
Final Answer: \[ \boxed{2700} \]