Q39 / 75 · JEE Main 2025 (28 Jan Shift 2) · Mathematics

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

  • A
    $x+9 y=36$
  • B
    $4 x-9 y=12$
  • Correct answer
    C
    $6 x-9 y=20$
  • D
    $9 x-9 y=32$

Solution

C: $x^2+y^2-8 x=0$ $\mathrm{H}: \frac{x^2}{9}-\frac{y^2}{4}=1$ By solving $\frac{x^2}{9}-\left(\frac{8 x-x^2}{4}\right)=1$ $\begin{aligned} & 4 x^2-72 x+9 x^2=36 \\ & \Rightarrow 13 x^2-72 x-36=0 \\ & \Rightarrow 13 x^2-78 x+6 x-36=0 \\ & \Rightarrow 13 x(x-6)+6(x-6)=0 \\ & \Rightarrow x=6 \text { or }-\frac{13}{6} \times \text { neglected } \\ & \Rightarrow y^2=8(6)-(6)^2 \\ & \Rightarrow y= \pm \sqrt{12} \end{aligned}$ So, points $A$ and $B$ are $(6, \sqrt{12}),(6,-\sqrt{12})$ $P\left(h, \frac{2 h+4}{3}\right)$ Centroid of $\triangle P A B$ is $\left(\frac{12+h}{3}, \frac{2 h+4}{9}\right)$ By options this centroid lies on the live $6 x-9 y=20$

Hence, the correct answer is option: 'C'.

More Hyperbola Questions