Equation of the locus of the centroid of the triangle whose vertices are $\left(\mathrm{a} \cos \mathrm{k}, \mathrm{a} \sin \mathrm{k}\right), \left(\mathrm{b} \sin \mathrm{k},-\mathrm{b} \cos \mathrm{k}\right)$ and $\left(1, 0\right),$( where $\mathrm{k}$ is  parameter) is

$$\begin{gathered} \mathrm{Let} \mathrm{P}(\mathrm{x},\mathrm{y}) \mathrm{be} \mathrm{locus} \mathrm{of} \mathrm{centriod} \mathrm{of} \mathrm{triangle} \mathrm{given} \mathrm{vertices} \mathrm{are} \\ \mathrm{Let} \mathrm{A}=\left(\mathrm{a} \cos \mathrm{k}, \mathrm{a} \sin \mathrm{k}\right), \mathrm{B}=(\mathrm{b} \sin \mathrm{k},-\mathrm{b} \cos \mathrm{k}), \mathrm{c}=(1,0) \\ \mathrm{centroid} \mathrm{of} {∆}^{le}\mathrm{ABC}=(\mathrm{x},\mathrm{y}) \\ \Rightarrow \left[\frac{\mathrm{a} \cos \mathrm{k}+\mathrm{b} \sin \mathrm{k}+1}{3},\frac{\mathrm{a} \sin \mathrm{k}-\mathrm{b} \cos \mathrm{k}}{3}\right]=\left[\mathrm{x},\mathrm{y}\right] \\ \Rightarrow \mathrm{a} \cos \mathrm{k}+\mathrm{b} \sin \mathrm{k}+1=3\mathrm{x}, \mathrm{a} \sin \mathrm{k}-\mathrm{b} \cos \mathrm{k}=3\mathrm{y} \\ \Rightarrow \mathrm{a} \cos \mathrm{k}+\mathrm{b} \sin \mathrm{k}=3\mathrm{x}-1\to \left(1\right), \mathrm{a} \sin \mathrm{k}-\mathrm{b} \cos \mathrm{k}=3\mathrm{y}\to \left(2\right) \\ {\left(1\right)}^2+{\left(2\right)}^2 \Rightarrow {(3\mathrm{x}-1)}^2+9{\mathrm{y}}^2={(\mathrm{a} \cos \mathrm{k}+\mathrm{b} \sin \mathrm{k})}^2+{(\mathrm{a} \sin \mathrm{k}-\mathrm{b} \cos \mathrm{k})}^2 \\ \Rightarrow (3\mathrm{x}-1{)}^2+9{\mathrm{y}}^2={\mathrm{a}}^2({\cos }^2\mathrm{k}+{\sin }^2\mathrm{k})+{\mathrm{b}}^2({\sin }^2+{\cos }^2\mathrm{k}) \\ \Rightarrow {(3\mathrm{x}-1)}^2+9{\mathrm{y}}^2={\mathrm{a}}^2+{\mathrm{b}}^2 \end{gathered}$$