JEE MAIN - Mathematics (2025 - 28th January Evening Shift - No. 12)

If A and B are the points of intersection of the circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ and a point P moves on the line $2x - 3y + 4 = 0$, then the centroid of $\Delta PAB$ lies on the line :
$x + 9y = 36$
$9x - 9y = 32$
$4x - 9y = 12$
$6x - 9y = 20$

Explanation

$$\begin{aligned} & x^2+y^2-8 x=0, \frac{x^2}{9}-\frac{y^2}{4}=1 \quad\text{.... (1)}\\ & 4 x^2-9 y^2=36 \quad\text{.... (2)}\\ & \text { Solve }(1) \&(2) \\ & 4 x^2-9\left(8 x-x^2\right)=36 \\ & 13 x^2-72 x-36=0 \\ & (13 x+6)(x=6)=0 \\ & x=\frac{-6}{13}, x=6 \\ & x=\frac{-6}{13}(\text { rejected }) \\ & y \rightarrow \text { Imaginary } \\ & n=6, \frac{36}{9}-\frac{y^2}{4}=1 \\ & y^2=12, y=I \sqrt{12} \\ & A(6, \sqrt{12}), B(6,-\sqrt{12}) \\ & p\left(\alpha, \frac{2 \alpha+4}{3}\right) P \text { lies on } \end{aligned}$$

$$\begin{aligned} & \text { centroid }(\mathrm{h}, \mathrm{k}) \quad 2x-3y+y=0\\ & \mathrm{h}=\frac{12+\alpha}{3}, \alpha=3 \mathrm{~h}-12 \\ & \mathrm{k}=\frac{\frac{2 \alpha-3 y}{3}}{3} \Rightarrow 2 \alpha+4=9 \mathrm{y} \\ & \alpha=\frac{9 \mathrm{k}-4}{2} \\ & 6 \mathrm{~h}-2 \mathrm{y}=9 \mathrm{k}-4 \\ & 6 \mathrm{x}-9 \mathrm{y}=20 \end{aligned}$$

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