The diagram (attached) shows a sketch of the curve with parametric equations 'x=t^3', 'y=3t^3 along with the normal at the point with parameter t=2.
Find the equation of this normal and the coordinates of this point where it cuts the curve again.
I found dy/dx to be 6t/(3t^2-12), and am stumped on where to go from here. Clearly the normal at t=2 has something to do with this but I can't presently see how. Any ideas guys?
Thanks for any help