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 $\bigtriangleup {PSS}'$ is equal to :
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Solution
$\ P(10,2\sqrt{15})$ lies on $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{{\text{ }b}^{2}} = 1$
$$\begin{array}{r} \therefore\frac{100}{a^{2}} - \frac{60}{{\text{ }b}^{2}} = 1\#(1) \end{array}$$
∵ length of latus rectum $= 8$
$$\begin{array}{r} \frac{2 \cdot {\text{ }b}^{2}}{a} = 8 \Rightarrow \frac{{\text{ }b}^{2}}{a} = 4\#(2) \end{array}$$
From (1) & (2)
$${\frac{100}{a^{2}} - \frac{60}{4a} = 1 }{400 - 60a = 4a^{2} }{4a^{2} + 60a - 400 = 0 }{a^{2} + 15a - 100 = 0 }$$$a = 5\& - 20$ (rejected)
$$\Rightarrow b = \sqrt{20} $$∴ Hyperbola is $\frac{x^{2}}{25} - \frac{y^{2}}{20} = 1$
∴ Focal length $S_{1}{\text{ }S}_{2} = 2ae = 2.5 \cdot \left( \sqrt{1 + \frac{4}{5}} \right) = 6\sqrt{5}$
∴ Area of $\Delta{PS}_{1}{\text{ }S}_{2} = \frac{1}{2} \cdot 6\sqrt{5} \cdot 2\sqrt{15} = 30\sqrt{3} = A$
$$\therefore A^{2} = 2700$$
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