The motion of a particle is given by $\frac{dx}{dt}=$

$-3a{\cos }^{2}t\sin t,\frac{dy}{dt}=3a{\sin }^{2}t\cos t$ and passes through at $t$

=0, then the path is given by

$\mid x=-3a\int {\cos }^{2}t\sin tdt+C=a{\cos }^{3}t+C$

for $t=0,C=-a$

$y=a\int {\sin }^{2}t\cos tdt=a{\sin }^{3}t+G,\text{ at }t=0,y=0\text{ so }{C}_1=0$

Thus $x=a\left({\cos }^{3}t-1\right),y=a{\sin }^{3}t$

$\begin{array}{l}\Rightarrow {\left(\frac{x}{a}+1\right)}^{1/3}=\cos t,{\left(\frac{y}{a}\right)}^{1/3}=\sin t\\ \Rightarrow {\left(\frac{x}{a}+1\right)}^{2/3}+{\left(\frac{y}{a}\right)}^{2/3}=1.\Rightarrow (x+a{)}^{2/3}+y^{2/3}=a^{2/3}\end{array}$