1. The problem statement, all variables and given/known data Hello, So i was just doing a few past papers for an upcoming exam and i got stuck on this one. i have got to the stage where i have 1/p all in term of x and y, but i can't see how you would go about getting the maximum acceleration just from x and y values. 2. Relevant equations a=v^2/r 3. The attempt at a solution so here is how i went about it: y^2 = 4ax therefore by implicit differentiation 2ydy/dx = 4a thus dy/dx = 2a/y for the equation for 1/p we need (dy/dx)^2 which is, 4a^2/y^2 →4a^2/4ax since y^2 = 4ax thus we get (dy/dx)^2 = a/x now we need d^2y/dx^2 so by implicit differentiation again 2a*y^-1 → -2a*y^-2*dy/dx (-2a/y^2)(2a/y) → (-2a/4ax)(2a/y) → -a/xy therefore 1/p = ((1+a/x)^(3/2))/(-a/xy) This is where I found it difficult to continue. I tried by saying a is a maximum when p is a minimum (a=v^2/p) so da/dp = -v^2/p^2, but i couldn't really get any further any help would be greatly appreciated because it is just annoying me now haha.