Let P be a point on the hyperbola in the first quadrant such that the area of triangle formed by P and the two foci of H is . Then, the square of the distance of P from the origin is [2024]

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 [2024]

Smart Achievers · Accepted Answer

(4)

We have, $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$

Here, $a^{2} = 9, b^{2} = 4$

$b^{2} = a^{2} (e^{2} - 1) \Rightarrow e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow e^{2} = 1 + \frac{4}{9} = \frac{13}{9}$

$\Rightarrow e = \frac{\sqrt{13}}{3} \Rightarrow s_{1} s_{2} = 2 a e = 2 \times 3 \times \frac{\sqrt{13}}{3} = 2 \sqrt{13}$

Area of $△ P S_{1} S_{2} = \frac{1}{2} \times β \times s_{1} s_{2} = 2 \sqrt{13}$

$\Rightarrow \frac{1}{2} \times β \times (2 \sqrt{13}) = 2 \sqrt{13} \Rightarrow β = 2$

$\frac{α^{2}}{9} - \frac{β^{2}}{4} = 1 \Rightarrow \frac{α^{2}}{9} = 2 \Rightarrow α^{2} = 18 \Rightarrow α = 3 \sqrt{2}$

Distance of P from origin = $\sqrt{α^{2} + β^{2}} = \sqrt{18 + 4} = \sqrt{22}$ .

Solution

Q.
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 [2024]

1 20

2 18

3 26

4 22

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