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 $ riangle \mathrm{PAB}$ lies on the line :

Question

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 $	riangle \mathrm{PAB}$ lies on the line :

Aicharya · Accepted Answer

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 $6 x-9 y=20$

If $A$ and $B$ are the points of intersection of the circle $x^2+y^2-8 x=0$ and the hyperbola…

Solution