$x = 2a{t^2},y = a{t^4}$

Here, $x = 2a{t^2}$ ....(i)
and $y = a{t^4}$ ...(ii)
Differentiating (i) \& (ii) w.r.t. t, we get

$\cfrac{{dx}}{{dt}} = 2a(2t) = 4at$ and $\cfrac{{dy}}{{dt}} = 4a{t^3}$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{{dy/dt}}{{dx/dt}} = \cfrac{{4a{t^3}}}{{4at}} = {t^2}$