Let the product of focal distances of the point on the hyperbola be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then is equal to __________. [2025]

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Solution

Q.
Let the product of focal distances of the point $P (4, 2 \sqrt{3})$ on the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be 32. Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then $p^{2} + q^{2}$ is equal to __________. [2025]

Ans.
(120)

We have, $P (4, 2 \sqrt{3})$

Now, $P S_{1} \cdot P S_{2} = 32$ ... (i)

where $S_{1} = (a e, 0) and S_{2} = (- a e, 0)$

ALso, $| P S_{1} - P S_{2} | = 2 a$

Now, $P (4, 2 \sqrt{3})$ lies on H

$∴ \frac{16}{a^{2}} - \frac{12}{b^{2}} = 1$

$\Rightarrow 16 b^{2} - 12 a^{2} = a^{2} b^{2}$ ... (ii)

Now, $| P S_{1} - P S_{2} |^{2} = 4 a^{2}$

$\Rightarrow P S_{1}^{2} + P S_{2}^{2} - 2 P S_{1} \cdot P S_{2} = 4 a^{2}$

$\Rightarrow {(a e - 4)}^{2} + 12 + {(a e + 4)}^{2} + 12 - 64 = 4 a^{2}$

$\Rightarrow 2 a^{2} e^{2} - 8 = 4 a^{2} \Rightarrow a^{2} + b^{2} - 4 = 2 a^{2}$

$\Rightarrow b^{2} - a^{2} = 4$ ... (iii)

From equation (ii) and (iii)

$\Rightarrow 16 (a^{2} + 4) - 12 a^{2} = a^{2} (a^{2} + 4)$

$\Rightarrow 16 a^{2} + 64 - 12 a^{2} = a^{4} + 4 a^{2}$

$\Rightarrow a^{4} = 64 \Rightarrow a^{2} = 8 ∴ b^{2} = 12$

Now, $p^{2} + q^{2} = 4 b^{2} + \frac{4 b^{4}}{a^{2}} = 120$ .

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