A path is described by the position vector $\mathbf{r}$: $$\mathbf{r}=a\cos(\omega t)\mathbf{\hat{i}}+b\sin{\omega t}\mathbf{\hat{j}}$$ I am asked to show that the path is the ellipse in the form of: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Is this not converting from polar to cartesian using $x=\mathbf{r}\cos(\omega t)$ and $y=\mathbf{r}\sin(\omega t)$? When I tried this I got as far as $x^2+y^2-ax-by=0$ and I am not sure what to do next, thanks.
Hint: Consider above $\bf{r}$ as $$\mathbf{r}=x(t)\mathbf{\hat{i}}+y(t)\mathbf{\hat{j}}$$ and then satisfy $x(t),~y(t)$ in the given Cartesian equation.