If `x=2(cost-sint)` and `y=3 (cost+ sint)` represents a conic,its foci are

If x=sint and y=cos pt, then

If x=sint and y=cos2t , then (dy)/(dx) =

If x=a(cost+tsint) and y=a(sint- tcost) , find (d^2y)/(dx^2) .

If x = sint cos2t and y = cost sin2t , then at t = pi/4 , the value of dy/dx is equal to:

If x=f(t)cost-f^(prime)(t)sint and y=f(t)sint+f^(prime)(t)cost , then ((dx)/(dt))^2+((dy)/(dt))^2= (a) f(t)-f^(primeprime)(t) (b) {f(t)-f^(primeprime)(t)}^2 (c) {f(t)+f^(primeprime)(t)}^2 (d) none of these

If x=a(cost+tsint) and y=a(sint-tcost) , find (d^(2)y)/(dx^(2)) .

If x=a(t-sint), y=a(1-cost) then find (d^2y)/(dx^2) .

If x=a(t+sint),y=a(1-cost)," then "(dy)/(dx)=

x=3cost-2cos^(3)t,y=3sint-2sin^(3)t