Let P(10,2√15) be a point on the hyperbola x^2a^2-y^2b^2=1, whose foci are S and S^'.

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 $\Delta{PSS}^{'}$ is equal to :

Accepted Answer

$\begin{gathered} P(10,2\sqrt{15})\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1 \ 	herefore \frac{100}{{a}^{2}}-\frac{60}{{b}^{2}}=1\ldots \ldots \ldots \ldots ..(1) \ \because =8 \ \frac{2\cdot {b}^{2}}{a}=8\Rightarrow \frac{{b}^{2}}{a}=4\ldots \ldots \ldots .(2) \end{gathered}$

From (1) \& (2)

$\begin{gathered} \frac{100}{{a}^{2}}-\frac{60}{4a}=1 \ 400-60a=4{a}^{2} \ 4{a}^{2}+60a-400=0 \ {a}^{2}+15a-100=0 \ a=5-20() \ \Rightarrow b=\sqrt{20} \end{gathered}$

$\begin{gathered} 	herefore \frac{{x}^{2}}{25}-\frac{{y}^{2}}{20}=1 \ 	herefore S{S}^{1}=2ae=2\cdot 5\cdot (\sqrt{1+\frac{4}{5}})=6\sqrt{5} \ 	herefore 	rianglePS{S}^{1}=\frac{1}{2}\cdot 6\sqrt{5}\cdot 2\sqrt{15}=30\sqrt{3}=A \ 	herefore {A}^{2}=2700 \end{gathered}$

Solutions · 1