The correct option is D
the acceleration of the particle is always directed towards a focusx=a cos pt ⇒ cos(pt)=xa
...(i) y=b sin pt ⇒ sin(pt)=yb
Squaring and adding (i) and (ii), we get x2a2+y2b2=1 ∴
Path of the particle is in ellipse.
Hence option (a) is correct.
From the given equations
We can find, dxdt=vx=−ap sin pt d2x=ax=−ap2 cos pt dydt=vy=bp cos pt
and d2ydt=ay=−bp2 sin pt
At time t=π2p or pt = π2 ax and vy
become zero (because cos π2=0) only vx and ay are left.
or we can say that velocity is along negative x-axis and acceleration along y-axis.
Hence, at t=π2p,
velocity and acceleration force the particle are normal to each other. So option (b) is also correct. At t = t, position of the particle →r(t)=x^i+y^j= a cos pt^i+b sin pt^j
and acceleration of the particle is →a(t)=ax^i+ay^j=p2[a cos pt^i+b sin pt^j] =p2[x^i+y^j=−p2→r(t)]
Therefore acceleration of the particle is always directed towards origin.
Hence option (c) is also correct.
At t = 0, particle is at (a, 0) and at t=π2p,
particle is at (0, b). Therefore, the distance covered is one fourth of the elliptical path not a.
Hence option (d) is wrong.