Question

show that the radius of the curvature of the point 3a/2,3a/2 on the curve x3+y3=3axy is -8√2/3a.​

Answer 1

Step-by-step explanation:

make the orthogonal change of coordinates x+y=2–√u, x−y=2–√v, then you can solve v=±f(u), where f(u) is some smooth function around u=0, f(0)=0 and f′(0)≠0. So our curve has a self-intersection at the origin, with each "branch" having the same curvature at the origin. Now take Taylor series at u=0, get v=±(Au+Bu2+...), which reduces to finding the curvature at the origin of the parabola v=Au+Bu2.