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

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Hyperbola

Let the product of the focal distances of the point

P (4, 23)

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

Answer: 120

Since

P (4, 23)

lies on

H

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

. The product of the focal distances at

P

∣ e^{2} x^{2} - a^{2} ∣

; with

e^{2} = \frac{a ^{2} + b ^{2}}{a ^{2}}

this is

16 + \frac{16 b ^{2}}{a ^{2}} - a^{2} = 32

, which rearranges to

16 b^{2} = a^{4} + 16 a^{2}

. Combined with the point condition

16 b^{2} - 12 a^{2} = a^{2} b^{2}

, these give

b^{2} = a^{2} + 4

and then

a^{4} = 64

, so

a^{2} = 8, b^{2} = 12

. The conjugate axis is

p = 2 b = 43

and the latus rectum is

q = \frac{2 b ^{2}}{a} = \frac{24}{2 2} = 62

. Therefore

p^{2} + q^{2} = 48 + 72 = 120

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Integer answer