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 sqrt[8] 5/3 . If for H , the length of the latus rectum is I and the product of the focal distance of the point P is m , then 9l 2+ 6 m is equal to:

Mathematics · Hyperbola

JEE Main 2025 — 4 April, Evening Shift — Question 34

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 $\sqrt[8]{\frac{5}{3}}$ . If for

$H$ , the length of the latus rectum is $I$ and the product of the focal distance of the point $P$ is $m$ , then $9l^2+$ 6 m is equal to:

Option A:
186
Option B:
187
Option C:
184
Option D:
185
Correct

Answer: D

Step-by-step solution

$\sqrt{(c+4)^{2}+9}+\sqrt{(c-4)^{2}+9}=8 \sqrt{\frac{5}{3}}$

Solving, $c=\frac{5}{\sqrt{6}}=\mathrm{al} \Rightarrow a^{2}\left(1+\frac{b^{2}}{a^{2}}\right)=\frac{25}{6} \Rightarrow a^{2}+b^{2}=\frac{25}{6}$

$\frac{16}{a^{2}}-\frac{9}{b^{2}}=1$

$16 b^{2}-9 d^{2}=a^{2} b^{2} \Rightarrow 16\left(\frac{25}{6}-a^{2}\right)-9 a^{2}=9 a^{2} b^{2}$

$P F_{1}+P F_{2}=8 \sqrt{\frac{5}{3}} \Rightarrow a^{2}=\frac{5}{2}, b^{2}=\frac{5}{3}$

$\left|P F_{1}-P F_{2}\right|=2 a$

$\frac{64.5}{3}=4 \mathrm{a}^{2}+4 m \Rightarrow m=\frac{80}{3}-a^{2}$

$6 m=160-6 a^{2}$

$9 \ell^{2}=9\left(\frac{2 b^{2}}{a}\right)^{2}=\frac{36 b^{4}}{a^{2}}$

$9 \ell^{2}+6 m=\frac{36\left(\frac{25}{9}\right)}{\frac{5}{2}}+160-6\left(\frac{5}{2}\right)$

$=\frac{72 \times 5}{9}+160-15$

$=160+40-15=185$

Answer key and solution verified before publishing.

Practise Hyperbola

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Hyperbola
Topic: Tangents & normals to hyperbola, chord of contact