JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Hyperbola

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Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be

(- 5, 0)

and

5 x + 9 = 0

, respectively. If the product of the focal distances of a point

(α, 25)

on the hyperbola is p , then 4 p is equal to ______.

Answer: 189

Solution

From the focus,

a e = 5

, and from the directrix

\frac{a}{e} = \frac{9}{5}

. Multiplying,

a^{2} = 9

, so

a = 3

and

e = \frac{5}{3}

, giving

b^{2} = a^{2} (e^{2} - 1) = 16

. The hyperbola is

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

; substituting

(α, 25)

gives

\frac{α ^{2}}{9} - \frac{20}{16} = 1

, so

α^{2} = \frac{81}{4}

. The product of the focal distances is

∣ e α - a ∣ ∣ e α + a ∣ = e^{2} α^{2} - a^{2} = \frac{25}{9} \cdot \frac{81}{4} - 9 = \frac{189}{4}

. Therefore

4 p = 189

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Integer answer

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JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Hyperbola

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