The resultant of two rectangular simple harmonic motion of the same frequency and unequal amplitudes but differing in phase by $\mathrm{\pi }/2$ is

If first equation is $\mathrm{x}={\mathrm{a}}_1\sin \mathrm{\omega t}\Rightarrow \sin \mathrm{\omega t}=\frac{\mathrm{x}}{{\mathrm{a}}_1}$

Then second equation will be $\mathrm{y}={\mathrm{a}}_2\sin \left(\mathrm{\omega t}+\frac{\mathrm{\pi }}{2}\right)$ $={\mathrm{a}}_2\left[\sin \mathrm{\omega tcos} \frac{\mathrm{\pi }}{2}+\cos \mathrm{\omega tsin} \frac{\mathrm{\pi }}{2}\right]={\mathrm{a}}_2\cos \mathrm{\omega t}\Rightarrow \cos \mathrm{\omega t}=\frac{\mathrm{y}}{{\mathrm{a}}_2}$

By squaring and adding Eqs. (1) and (2) 

${\sin }^2 \mathrm{\omega t}+{\cos }^2 \mathrm{\omega t}=\frac{{\mathrm{x}}^2}{{\mathrm{a}}_1^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}_2^2}$

$\Rightarrow \frac{{\mathrm{x}}^2}{{\mathrm{a}}_1^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}_2^2} =1$

This is equation of an ellipse.