JEE Main 2025 Apr 4 Shift 2, Mathematics Q10: Introduction to Hyperbola

JEE Main2025Apr 4, Shift 2Mathematics

Let the sum of the focal distances of the point $P (4, 3)$ on the hyperbola $H : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be $8 \frac{5}{3}$ . 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 :-

A
184
B
186
C
185
D
187

Solution

For a hyperbola, the focal distances of $P (x_{0}, y_{0})$ are $∣ e x_{0} \pm a ∣$ . Their sum (for a point on the right branch with $x_{0} = 4$ ) is $2 e x_{0} = 8 e$ . Thus

$8 e = 8 \frac{5}{3} \Rightarrow e^{2} = \frac{5}{3}$ .

Using $b^{2} = a^{2} (e^{2} - 1)$ : $b^{2} = a^{2} \cdot \frac{2}{3}$ .

Since $P (4, 3)$ lies on $H$ : $\frac{16}{a ^{2}} - \frac{9}{b ^{2}} = 1$ . Substitute $b^{2} = \frac{2 a ^{2}}{3}$ :

$\frac{16}{a ^{2}} - \frac{27}{2 a ^{2}} = 1 \Rightarrow \frac{32 - 27}{2 a ^{2}} = 1 \Rightarrow a^{2} = \frac{5}{2}, b^{2} = \frac{5}{3}$ .

Latus rectum: $l = \frac{2 b ^{2}}{a}$ , so $l^{2} = \frac{4 b ^{4}}{a ^{2}} = \frac{4 \cdot 25/9}{5/2} = \frac{40}{9}$ . Hence $9 l^{2} = 40$ .

Product of focal distances: $m = (e x_{0} + a) (e x_{0} - a) = e^{2} x_{0}^{2} - a^{2} = \frac{5}{3} \cdot 16 - \frac{5}{2} = \frac{80}{3} - \frac{5}{2} = \frac{145}{6}$ . So $6 m = 145$ .

$∴ 9 l^{2} + 6 m = 40 + 145 = 185$ .

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JEE Main 2025 Apr 4 Shift 2, Mathematics Q10: Introduction to Hyperbola

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