Mathematics - Hyperbola Question with Solution | TestHub

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