Given, vertices of a triangle are
$A(a \cos \theta, a \sin \theta), B(b \sin \theta, b \cos \theta)$ and $C(1,0)$
$\therefore$ Let the locus of centroid be $(x, y)$.
$$
\begin{aligned}
& \therefore(x, y)=\left(\frac{a \cos \theta+b \sin \theta+1}{3}, \frac{b \cos \theta+0}{3}\right) \\
& \Rightarrow x=\frac{a \cos \theta+b \sin \theta+1}{3} \\
& \text { and } \quad y=\frac{a \sin \theta-b \cos \theta}{3}
\end{aligned}
$$
and $\quad y=\frac{a \sin \theta-b \cos \theta}{3}$
$$
\Rightarrow a \cos \theta+b \sin \theta=x-1
$$
and $\quad a \sin \theta-b \cos \theta=3 y$
$$
\Rightarrow a^2 \cos ^2 \theta+b^2 \cos ^2 \theta+2 a b \sin \theta \cos \theta=(3 x-1)^2
$$
and $a^2 \sin ^2 \theta+b^2 \cos ^2 \theta-2 a b \sin \theta \cos \theta=9 y^2$
On adding, we get
$$
\begin{aligned}
& a^2\left(\sin ^2 \theta+\cos ^2 \theta\right)+b^2\left(\cos ^2 \theta+\sin ^2 \theta\right) \\
&=(3 x-1)^2+9 y^2 \\
& \Rightarrow \quad a^2+b^2=(3 x-1)^2+9 y^2
\end{aligned}
$$