Question

Find the radius of curvature for the curve x³ + y³ = 3axy on the point (3a/2, 3a/2).

Answer 1

According to question

x³ + y³ = 3axy

Thus at point (\(\frac{3a}{2}\), \(\frac{3a}{2}\))

Putting the coordinates in the equation

\(\frac{27a^3}{8}\) + \(\frac{27a^3}{8}\) = \(\frac{54a^3}{4}\)

\(\frac{27a^3 + 27a^3}{8}\) = \(\frac{54a^3}{4}\)

\(\frac{54a^3}{8}\) = \(\frac{54a^3}{4}\)

Thus

a = \(\frac{1}{2}\)

Thus the coordinates are (\(\frac{1}{2}\), \(\frac{1}{2}\)).

Answer 2

Radius of curvatur=-3a/8√2