Hyperbola Question - None

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 the origin is

Accepted Answer

Given Equation for hyperbola is;
$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$
$\Rightarrow a^{2}=9, b^{2}=4$
We know that
$b^{2}=a^{2}\left(e^{2}-1 ight) \Rightarrow e^{2}=1+\frac{b^{2}}{a^{2}}$
$e^{2}=1+\frac{4}{9}=\frac{13}{9}$
$e=\frac{\sqrt{13}}{3} \Rightarrow s_{1} s_{2}=2 a e=2 \sqrt{13}$
Area of $ riangle {PS}_{1} {~S}_{2}=\frac{1}{2} imes \beta imes {S}_{1} {~S}_{2}=2 \sqrt{13}$
$\Rightarrow \frac{1}{2} imes \beta imes(2 \sqrt{13})=2 \sqrt{13} \Rightarrow \beta=2$
$\frac{\alpha^{2}}{9}-\frac{\beta^{2}}{4}=1 \Rightarrow \frac{\alpha^{2}}{9}-1=1$
$\Rightarrow \alpha=3 \sqrt{2}$
Hence, Distance of $P$ from origin
$=\sqrt{\alpha^{2}+\beta^{2}}=\sqrt{22}$

Question:

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