Hyperbola

Concept:For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, latus rectum length is $\frac{2b^2}{a}$. The foci are at $(\pm ae,0)$ where $e = \sqrt{1+\frac{b^2}{a^2}}$.Explanation:Point $P(10, 2\sqrt{15})$ lies on the hyperbola. So $\frac{100}{a^2} - \frac{60}{b^2} = 1$. Given latus rectum $= 8$, so $\frac{2b^2}{a} = 8$ ⇒ $b^2 = 4a$. Substitute $b^2 = 4a$ into the first equation: $\frac{100}{a^2} - \frac{60}{4a} = 1$ ⇒ $\frac{100}{a^2} - \frac{15}{a} = 1$. Multiply by $a^2$: $100 - 15a = a^2$ ⇒ $a^2 + 15a - 100 = 0$. Solve: $(a+20)(a-5)=0$ ⇒ $a = 5$ (positive). Then $b^2 = 4 \times 5 = 20$ ⇒ $b = \sqrt{20}$. So hyperbola: $\frac{x^2}{25} - \frac{y^2}{20} = 1$. Eccentricity $e = \sqrt{1+\frac{20}{25}} = \sqrt{\frac{45}{25}} = \frac{3\sqrt{5}}{5}$. Distance between foci $SS' = 2ae = 2 \times 5 \times \frac{3\sqrt{5}}{5} = 6\sqrt{5}$. Area of $\triangle PSS'$ = $\frac{1}{2} \times SS' \times |y_P| = \frac{1}{2} \times 6\sqrt{5} \times 2\sqrt{15} = 30\sqrt{3}$. Square of area = $(30\sqrt{3})^2 = 900 \times 3 = 2700$.Answer:$2700$ → Option D.