Let P be a point on the hyperbola [H: (x^2)/(9)-(y^2)/(4)=1], in the first quadrant such t…

Question

Let $P$ be a point on the hyperbola $$H: \frac{x^2}{9}-\frac{y^2}{4}=1$$, in the first quadrant such that the area of triangle formed by $P$ and the two foci of $H$ is $$2 \sqrt{13}$$. Then, the square of the distance of $P$ from th...

Accepted Answer

$$\begin{aligned}
& \frac{x^2}{9}-\frac{y^2}{4}=1 \
& a^2=9, b^2=4 \
& b^2=a^2\left(e^2-1\right) \Rightarrow e^2=1+\frac{b^2}{a^2} \
& e^2=1+\frac{4}{9}=\frac{13}{9} \
& e=\frac{\sqrt{13}}{3} \Rightarrow s_1 s_2=2 \mathrm{ae}=2 \times 3 \times \sqrt{\frac{13}{3}}=2 \sqrt{13}
\end{aligned}$$

$$\begin{aligned}
& \text { Area of } \triangle \mathrm{PS}_1 \mathrm{~S}_2=\frac{1}{2} \times \beta \times \mathrm{S}_1 \mathrm{~S}_2=2 \sqrt{13} \
& \Rightarrow \frac{1}{2} \times \beta \times(2 \sqrt{13})=2 \sqrt{13} \Rightarrow \beta=2 \
& \begin{aligned}
& \frac{\alpha^2}{9}-\frac{\beta^2}{4}=1 \Rightarrow \frac{\alpha^2}{9}-1=1 \Rightarrow \alpha^2=18 \Rightarrow \alpha=3 \sqrt{2} \
& \text { Distance of P from origin }=\sqrt{\alpha^2+\beta^2} \
&=\sqrt{18+4}=\sqrt{22}
\end{aligned}
\end{aligned}$$

Solutions · 1