A particle which is simultaneously subjected to two perpendicular simple harmonic motions represented by; $$x = a_1 \cos \omega t$$ and $$y = a_2 \cos 2\omega t$$ traces a curve given by:

$$x = a_1 \cos \omega t \implies \cos \omega t = \frac{x}{a_1}$$

$$y = a_2 \cos 2\omega t$$

$$y = a_2 (2 \cos^2 \omega t - 1)$$

$$y = a_2 \left[ 2 \left( \frac{x}{a_1} \right)^2 - 1 \right]$$

$$y = \frac{2a_2}{a_1^2} x^2 - a_2$$

This equation is of the form $$y = kx^2 - c$$, which represents a parabola symmetric about the $$y$$-axis with its vertex at $$(0, -a_2)$$.

The curve opens upwards because the coefficient of $$x^2$$ is positive. At $$x = \pm a_1$$, $$y = a_2 [2(1) - 1] = a_2$$. At $$x = 0$$, $$y = -a_2$$.

This matches the plot shown in option D.