JEE Main 2025 — Mathematics Coordinate Geometry

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \quad....(1)$$a2x2​−b2y2​=1....(1)

$$\begin{aligned} & \mathrm{P}(4,2 \sqrt{3}) \\ & \mathrm{PS}_1 \cdot \mathrm{PS}_2=32 \\ & \left|\mathrm{PS}_1-\mathrm{PS}_2\right|=2 \mathrm{a} \\ & \mathrm{P}(4,2 \sqrt{3}) \text { lies on } \mathrm{H} \\ & \therefore \frac{16}{\mathrm{a}^2}-\frac{12}{\mathrm{~b}^2}=1\end{aligned}$$​P(4,23​)PS1​⋅PS2​=32∣PS1​−PS2​∣=2aP(4,23​) lies on H∴a216​− b212​=1​

$$16 b^2-12 a^2=a^2 b^2 \quad....(2)$$16b2−12a2=a2b2....(2)

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

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

$$\begin{aligned} & \therefore \mathrm{b}^2=12 \\ & \mathrm{p}^2+\mathrm{q}^2=4 \mathrm{~b}^2+\frac{4 \mathrm{~b}^4}{\mathrm{a}^2} \\ & =120\end{aligned}$$​∴b2=12p2+q2=4 b2+a24 b4​=120​