Answer
$\left(\frac{x}{5}\right)^{2}+\left(\frac{y}{2}\right)^{2}=1$
from the point $(0,-2)$ clockwise around the ellipse $3$ times to the point $(0,-2)$.
Work Step by Step
$x$ = $5\sin t$
$\sin t$ = $\frac{x}{5}$
$y$ = $2\cos t$
$\cos t$ = $\frac{y}{2}$
Use the Pythagorean Identity:
$\cos^2 z+\sin^2 z=1$
$\sin^{2}t+\cos^{2}t$ = $\left(\frac{x}{5}\right)^{2}+\left(\frac{y}{2}\right)^{2}$ = $1$
The motion of the particle takes place on an ellipse centered at $(0,0)$. As $t$ goes from $-\pi$ to $5\pi$, the particle starts at the point $(0,-2)$ and moves clockwise around the ellipse $3$ times because we have:
$t=-\pi\Rightarrow (0,-2)$
$t=0\Rightarrow (0,2)$
$t=\pi\Rightarrow (0,-2)$
$t=2\pi\Rightarrow (0,2)$
$t=3\pi\Rightarrow (0,-2)$
$t=4\pi\Rightarrow (0,2)$
$t=5\pi\Rightarrow (0,-2)$
