If 3ay^2=x(x-a)^2 with a>0, prove that the

Question

radius of curvature at the point (3a, 2a) is
50a/3.
Help with working plz

Answer 1

Just use the formula:

r = (1+y'^2)^(3/2) / |y"|

3ay^2 = x^3-2ax^2+ax
6ay y' = 3x^2-4ax+a
y' = (3x^2-4ax+a)/(6ay)
y" = [(6ay)(6x-4a) - (3x^2-4ax+a)(6ay')]/(36a^2y^2)

Now just plug in (3a,2a) and crank it out. Messy algebra, but not difficult.