This is an ellipse. Suppose for simplicity that the phase shift is $\pi/2$: $$ x(t)=A\cos(\omega t + \phi),\\ y(t)=B\sin(\omega t +\phi), $$ then $$ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1, $$ which is an equation for an ellipse. I suggest doing the case of unequal phases as an exercise.

This is an ellipse. Suppose for simplicity that the phase shift is $\pi/2$: $$ x(t)=A\cos(\omega t + \phi),\\ y(t)=B\sin(\omega t +\phi), $$ then $$ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1, $$ which is an equation for an ellipse. I suggest doing the case of unequal phases as an exercise (hint: express $\sin(\omega t),\cos(\omega t)$ in terms of $x,y$, and employ again the main trigonometric identity.)