Let P be a point on the hyperbola H: x ^ 2 9 - y ^ 2 4 = 1 , in the...

Plotting the diagram of given data we get,

Given equation of hyperbola is $\dfrac{{x}^{2}}{9} - \dfrac{{y}^{2}}{4} = 1$.

$⇒ {a}^{2} = 9, {b}^{2} = 4$

We know that, ${b}^{2} = {a}^{2}\left({e}^{2} - 1\right)$

$⇒ {e}^{2} = 1 + \dfrac{{b}^{2}}{{a}^{2}}$

$⇒ {e}^{2} = 1 + \dfrac{4}{9}$

$⇒ {e}^{2} = \dfrac{13}{9}$

$⇒ e = \dfrac{\sqrt{13}}{3}$

Now, ${S}_{1}{S}_{2} = 2ae$

$⇒ {S}_{1}{S}_{2} = 2 \times 3 \times \sqrt{\dfrac{13}{3}}$

$⇒ {S}_{1}{S}_{2} = 2\sqrt{13}$

It is given that, area of $\DeltaP{S}_{1}{S}_{2}$ is $2\sqrt{13}$.

$⇒ \dfrac{1}{2} \times \beta \times {S}_{1}{S}_{2} = 2\sqrt{13}$

$⇒ \dfrac{1}{2} \times \beta \times 2\sqrt{13} = 2\sqrt{13}$

$⇒ \beta = 2$

Since, $\left(\alpha, \beta\right)$ lies on the hyperbola,

$⇒ \dfrac{{\alpha}^{2}}{9} - \dfrac{{\beta}^{2}}{4} = 1$

$⇒ \dfrac{{\alpha}^{2}}{9} - 1 = 1$

$⇒ {\alpha}^{2} = 18$

$⇒ \alpha = 3\sqrt{2}$

Distance of $P$ from origin is given by,

$OP = \sqrt{{\alpha}^{2} + {\beta}^{2}}$

$⇒ OP = \sqrt{18 + 4}$

$⇒ OP = \sqrt{22}$

$⇒ O{P}^{2} = 22$