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
Your Idea is sound first you need to take the derivative
So the slope of the normals line is perpendicular to the tangent this gives
So the equation of the normal line is horizontal and since So the equation of the normal line
Now we just need to find another value of such that
Can you finish from here?