c
From the two mutually perpendicular $S.H.M. 's$, the general equation of Lissajous figure
$\frac{{{x^2}}}{{{A^2}}} + \frac{{{y^2}}}{{{B^2}}} - \frac{{2xy}}{{AB}}\cos \,\delta = {\sin ^2}\,\delta$
$x = A\,\sin \,\left( {at + \delta } \right)$
$y = B\,\sin \,\left( {bt + r} \right)$
Clearly $A\ne B$ hence ellipse