JEE Main 2025 Apr 7 Shift 2, Mathematics Q24: Introduction to Hyperbola

JEE Main2025Apr 7, Shift 2Mathematics

Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $2 a$ and $2 b$ , 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 _____.

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Solution

The focus is at $(- c, 0) = (- 5, 0)$ , so $c = 5$ . The corresponding directrix is $x = - \frac{a}{e} = - \frac{9}{5}$ , so $\frac{a}{e} = \frac{9}{5}$ .

Also $c = a e = 5$ , so $a = \frac{5}{e}$ . Substituting into $\frac{a}{e} = \frac{9}{5}$ : $\frac{5}{e ^{2}} = \frac{9}{5} \Rightarrow e^{2} = \frac{25}{9} \Rightarrow e = \frac{5}{3}$ .

Then $a = \frac{5}{5/3} = 3$ , and $b^{2} = c^{2} - a^{2} = 25 - 9 = 16$ .

The hyperbola is $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ . Substituting the point $(α, 25)$ : $\frac{α ^{2}}{9} - \frac{20}{16} = 1 \Rightarrow α^{2} = 9 (1 + \frac{5}{4}) = \frac{81}{4}$ .

The product of focal distances is $P F_{1} \cdot P F_{2} = (e α - a) (e α + a) = e^{2} α^{2} - a^{2}$ .

$p = \frac{25}{9} \times \frac{81}{4} - 9 = \frac{225}{4} - 9 = \frac{189}{4}$ .

$∴ 4 p = 189$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q24: Introduction to Hyperbola

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