Radius of curvature of x³+y³=3axy at origin

For a curve passing through the origin, the formula of Newton is used: if the x-axis is the tangent at the origin, then ρ = lim_{x→0} x²/(2y); if the y-axis is the tangent, ρ = lim y²/(2x).

The curve x³ + y³ = 3axy has the lowest-degree terms 3axy = 0 at the origin, so the tangents at the origin are x = 0 and y = 0 (a node).

Tangent y = 0 (x-axis): near the origin y is small, so y³ is negligible compared with the other terms. Then x³ ≈ 3axy, giving y ≈ x²/(3a). Check: let y = c₂x² + ..., then x³ − 3a c₂ x³ = 0 gives c₂ = 1/(3a).

ρ = lim x²/(2y) = x²/(2x²/(3a)) = 3a/2

(Equivalently ρ = (1+y′²)^(3/2)/y″ with y′ = 0, y″ = 2/(3a).)

Tangent x = 0 (y-axis): by symmetry in x and y, x ≈ y²/(3a), and ρ = lim y²/(2x) = 3a/2.

Answer: the radius of curvature at the origin is 3a/2 (for both branches).