Hyperbola Question 5
Question 5 - 2024 (30 Jan Shift 2)
Let $\mathrm{P}$ be a point on the hyperbola $\mathrm{H}: \frac{\mathrm{x}^{2}}{9}-\frac{\mathrm{y}^{2}}{4}=1$, in the first quadrant such that the area of triangle formed by $\mathrm{P}$ and the two foci of $\mathrm{H}$ is $2 \sqrt{13}$. Then, the square of the distance of $\mathrm{P}$ from the origin is
(1) 18
(2) 26
(3) 22
(4) 20
Show Answer
Answer (3)
Solution
$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$
$a^{2}=9, b^{2}=4$
$b^{2}=a^{2}\left(e^{2}-1\right) \Rightarrow e^{2}=1+\frac{b^{2}}{a^{2}}$
$e^{2}=1+\frac{4}{9}=\frac{13}{9}$
$e=\frac{\sqrt{13}}{3} \Rightarrow s_{1} s_{2}=2 a e=2 \times 3 \times \sqrt{\frac{13}{3}}=2 \sqrt{13}$
Area of $\Delta \mathrm{PS}{1} \mathrm{~S}{2}=\frac{1}{2} \times \beta \times \mathrm{s}{1} \mathrm{~S}{2}=2 \sqrt{13}$
$\Rightarrow \frac{1}{2} \times \beta \times(2 \sqrt{13})=2 \sqrt{13} \Rightarrow \beta=2$
$\frac{\alpha^{2}}{9}-\frac{\beta^{2}}{4}=1 \Rightarrow \frac{\alpha^{2}}{9}-1=1 \Rightarrow \alpha^{2}=18 \Rightarrow \alpha=3 \sqrt{2}$
Distance of $\mathrm{P}$ from origin $=\sqrt{\alpha^{2}+\beta^{2}}$
$$ =\sqrt{18+4}=\sqrt{22} $$
