Question

The position of a particle moving in the $x-y$ plane varies with time $t$ as $\left(\frac{dx}{dt}\right)=8\pi \sin 2\pi t$ and $\left(\frac{dy}{dt}\right)=5\pi \cos 2\pi t$. When $t=0,x=8,y=0$. The path of the particle is

• A. a straight line
• B. a circle
• C. an ellipse
• D. a parabola

Answer 1

The correct option is C an ellipse

$dx=8\pi \sin 2\pi tdt$, on integration we get, $x=-4\cos 2\pi t+c$
$Att=0,x=8soc=12\frac{(x-12)}{4}=-\cos 2\pi t$.......................................(1)

Similarly, $y=\frac{5}{2}\sin 2\pi t+c$

$Att=0,y=0soc=0\frac{y}{\frac{5}{2}}=\sin 2\pi t$...........................................(2)

$\frac{{(x-12)}^2}{16}+\frac{4y^2}{25}=1\Leftarrow Equationofellipse$