The velocity of a particle moving in the $x-y$ plane is given by $\frac{dx}{dt}=8\pi \sin 2\pi t$ and $\frac{dy}{dt}=5\pi \cos 2\pi t$ where $t=0,x=8$ and $y=0$. The path of the particle is

The correct option is B An ellipse

Given:
$\frac{dx}{dt}=8\pi \sin 2\pi t\dots \left(i\right)$
$\frac{dy}{dt}=5\pi \cos 2\pi t\dots \left(ii\right)$
From equation $\left(i\right)$ we get,
$dx=8\pi \sin 2\pi tdt$
${\int }_8^{x}dx=8\pi {\int }_o^t\sin 2\pi tdt$
$x-8=8\pi .{\left[\frac{-\cos 2\pi t}{2\pi }\right]}_0^t$
$x=8-4(\cos 2\pi t-\cos 0)$
$x=12-4\cos 2\pi t\dots \left(iii\right)$
From equation $\left(ii\right)$ we get,
${\int }_0^{y}dy=5\pi {\int }_o^t\cos 2\pi tdt$
$y={\left[\frac{5\pi \sin 2\pi t}{2\pi }\right]}_0^t$
$y=\frac{5}{2}\sin 2\pi t\dots \left(iv\right)$
Hence,
from $\left(iii\right)$ $\cos 2\pi t=\frac{12-x}{4}$
from $\left(iv\right)$ $\sin 2\pi t=\frac{2y}{5}$
Therefore, Squaring and adding, equation of ellipse is obtained.