Let P be a point on the hyperbola H: x^2/9 - y^2/4 = 1, in the first quadrant such that

Question

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

Answer 1

Correct option is (3) 22

\(\frac{\mathrm{x}^{2}}{9}-\frac{\mathrm{y}^{2}}{4}=1\)

\(\mathrm{a}^{2}=9, \mathrm{~b}^{2}=4\)

\(b^{2}=a^{2}\left(e^{2}-1\right) \Rightarrow e^{2}=1+\frac{b^{2}}{a^{2}}\)

\(\mathrm{e}^{2}=1+\frac{4}{9}=\frac{13}{9}\)

\(\mathrm{e}=\frac{\sqrt{13}}{3} \Rightarrow \mathrm{s}_{1} \mathrm{~s}_{2}=2 \mathrm{ae}=2 \times 3 \times \sqrt{\frac{13}{3}}=2 \sqrt{13}\)

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 P from origin \(=\sqrt{\alpha^{2}+\beta^{2}}\)

\(=\sqrt{18+4}\)

\(=\sqrt{22}\)