Let the sum of the focal distances of the point \(\mathrm{P}(4,3)\) on the hyperbola \(\mathrm{H}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\) be \(8 \sqrt{\frac{5}{3}}\). If for H, the length of the latus rectum is \(\ell\) and the product of the focal distances of the point P is m, then \(9 \ell^{2}+6 \mathrm{~m}\) is equal to :-

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Hyperbola

Let the sum of the focal distances of the point

P (4, 3)

on the hyperbola

H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

8 \frac{5}{3}

. If for H, the length of the latus rectum is

ℓ

and the product of the focal distances of the point P is m, then

9 ℓ^{2} + 6 m

is equal to :-

Answer: (C)

For a point on the right branch the focal distances sum to

2 e x

, so

2 e (4) = 8 5/3

gives

e = 5/3

and

b^{2} = a^{2} (e^{2} - 1) = \frac{2}{3} a^{2}

. Substituting

P (4, 3)

into the hyperbola,

\frac{16}{a ^{2}} - \frac{9}{2 a ^{2} /3} = 1

yields

a^{2} = \frac{5}{2}, b^{2} = \frac{5}{3}

. The latus rectum

ℓ = \frac{2 b ^{2}}{a}

gives

9 ℓ^{2} = \frac{36 b ^{4}}{a ^{2}} = 40

, while the product of focal distances

m = e^{2} x^{2} - a^{2} = \frac{5}{3} \cdot 16 - \frac{5}{2} = \frac{145}{6}

gives

6 m = 145

. Hence

9 ℓ^{2} + 6 m = 40 + 145 = 185

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option