Let P be a point on the hyperbola H: frac{x^2 | vedclass

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(C) Given the hyperbola equation: $\frac{x^2}{9}-\frac{y^2}{4}=1$.Here,$a^2=9$ and $b^2=4$.The eccentricity $e$ is given by $b^2=a^2(e^2-1)$,so $e^2=1

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$22$

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(C) Given the hyperbola equation: $\frac{x^2}{9}-\frac{y^2}{4}=1$.Here,$a^2=9$ and $b^2=4$.The eccentricity $e$ is given by $b^2=a^2(e^2-1)$,so $e^2=1+\frac{b^2}{a^2} = 1+\frac{4}{9} = \frac{13}{9}$.Thus,$e=\frac{\sqrt{13}}{3}$.The distance between the two foci $S_1$ and $S_2$ is $2ae = 2 	imes 3 	imes \frac{\sqrt{13}}{3} = 2\sqrt{13}$.Let $P = (\alpha, \beta)$. The area of $\Delta PS_1S_2 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2ae) 	imes \beta = 2\sqrt{13}$.Substituting $2ae = 2\sqrt{13}$,we get $\frac{1}{2} 	imes (2\sqrt{13}) 	imes \beta = 2\sqrt{13}$,which implies $\beta=2$.Since $P$ lies on the hyperbola,$\frac{\alpha^2}{9}-\frac{\beta^2}{4}=1$. Substituting $\beta=2$,we get $\frac{\alpha^2}{9}-\frac{4}{4}=1$ $\Rightarrow \frac{\alpha^2}{9}=2$ $\Rightarrow \alpha^2=18$.The square of the distance of $P$ from the origin is $OP^2 = \alpha^2+\beta^2 = 18 + 2^2 = 18+4 = 22$.

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 the triangle formed by $P$ and the two foci of $H$ is $2 \sqrt{13}$. Then,the square of the distance of $P$ from the origin is

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