Let the product of the focal distances of the point P(4,2 3 ) on the hyperbola H: x^2 a^2 - y^2 ~b^2 =1 be 32 . Let the length of the conjugate axis of H be p and the length of its latus rectum be q . Then p^2+q^2 is

Correct answer: 0 (x^2)/(a^2)-(y^2)/(b^2)=1 ....(1) aligned & P(4,2 3 ) & PS_1 · PS_2=32 & |PS_1-PS_2|=2 a & P(4,2 3 ) lies on H & 16 a^2 - 12 b^2 =1 aligned 16 b^2-12 a^2=a^2 b^2 ....(2) aligned & |PS_1-PS_2|^2=4 a^2 & PS_1^2+PS_2 ^2-2 PS_1 · PS_2=4 a^2 & (ae-4)^2+12+(ae+4)^2+12-64=4 a^2 & 2 a^2 e^2-8=4 a^2 & a^2+b^2-4=2 a^2 & b^2-a^2=4 aligned aligned & (2) &(3) 16(a^2+4)-12 a^2=a^2(a^2+4) & 16 a^2+64-12 a^2=a^4+4 a^2 & a^4=64 & a^2=8 aligned aligned & b^2=12 & p^2+q^2=4 b^2+ 4 b^4 a^2 & =120 aligned

JEE Main 20253 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Hyperbola

JEE Main 2025 · 3 April · Morning Shift · Mathematics · Hyperbola

Let the product of the focal distances of the point P(4,2 3 ) on...

Let the product of the focal distances of the point

\mathrm{P}(4,2 \sqrt{3})

on the hyperbola

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

be 32 .
Let the length of the conjugate axis of

H

p

and the length of its latus rectum be q . Then

\mathrm{p}^2+\mathrm{q}^2

is equal to _______

Correct answer

Step-by-step solution

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \quad....(1)

\begin{aligned} & \mathrm{P}(4,2 \sqrt{3}) \\ & \mathrm{PS}_1 \cdot \mathrm{PS}_2=32 \\ & \left|\mathrm{PS}_1-\mathrm{PS}_2\right|=2 \mathrm{a} \\ & \mathrm{P}(4,2 \sqrt{3}) \text{ lies on } \mathrm{H} \\ & \therefore \frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{ b}^2}=1\end{aligned}

16 b^2-12 a^2=a^2 b^2 \quad....(2)

\begin{aligned}& \left|\mathrm{PS}_1-\mathrm{PS}_2\right|^2=4 \mathrm{a}^2 \\ & \mathrm{PS}_1^2+\mathrm{PS}_2{ }^2-2 \mathrm{PS}_1 \cdot \mathrm{PS}_2=4 \mathrm{a}^2 \\ & (\mathrm{ae}-4)^2+12+(\mathrm{ae}+4)^2+12-64=4 \mathrm{a}^2 \\ & 2 \mathrm{a}^2 \mathrm{e}^2-8=4 \mathrm{a}^2 \\ & \mathrm{a}^2+\mathrm{b}^2-4=2 \mathrm{a}^2 \\ & \mathrm{ b}^2-\mathrm{a}^2=4\end{aligned}

\begin{aligned}& (2) &(3) \Rightarrow 16\left(a^2+4\right)-12 a^2=a^2\left(a^2+4\right) \\ & \Rightarrow 16 a^2+64-12 a^2=a^4+4 a^2 \\ & \Rightarrow a^4=64 \\ & \Rightarrow a^2=8\end{aligned}

\begin{aligned} & \therefore \mathrm{b}^2=12 \\ & \mathrm{p}^2+\mathrm{q}^2=4 \mathrm{ b}^2+\frac{4 \mathrm{ b}^4}{\mathrm{a}^2} \\ & =120\end{aligned}

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