A) a straight line
B) an ellipse
C) a circle
D) a parabola
Correct Answer: B
Solution :
y-x graph gives the shape of path of particle \[\frac{dX}{dt}=8\pi \sin 2\pi t\] \[\int_{8}^{X}{dX}=\int_{0}^{t}{8\pi }\sin 2\pi \,t\,dt\] \[x-8=-\frac{8\pi }{2\pi }[\cos 2\pi ]_{0}^{t}\] \[x-8=4(1-\cos 2\pi t)\] \[\cos 2\pi t=\frac{x-12}{4}\] \[\frac{dy}{dt}=-5\pi \cos \,2\pi t\] \[\int_{0}^{y}{dy}=5\pi t\int_{0}^{t}{\cos 2\pi t}\] \[y=\frac{5}{2}\sin 2\pi t\] \[{{\left( \frac{x-12}{4} \right)}^{2}}+\frac{{{y}^{2}}}{{{\left( \frac{5}{2} \right)}^{2}}}=1\] \[\frac{{{(x-12)}^{2}}}{{{(4)}^{2}}}+\frac{{{y}^{2}}}{\frac{5}{{{(2)}^{2}}}}=1\]
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