If $\theta$ is parameter, $A=(a\cos \theta ,a\sin \theta )$ and $B=(b\sin \theta ,-b\cos \theta )$ $C=(1,0)$ then the locus of the centroid of $△ABC$ is 

$A=(a\cos \theta ,a\sin \theta ),B=(b\sin \theta ,-b\cos \theta ),C=(1,0)$

Let $G=(x,y)$ 

$\begin{array}{c}(x,y)=(\frac{a\cos \theta +b\sin \theta +1}{3},\frac{a\sin \theta -b\sin \theta }{3})\\ 3x-1=a\cos \theta +b\sin \theta \\ 3y=a\sin \theta -b\cos \theta \\ {(3x-1)}^2+9y^2=a^2+b^2\end{array}$