This curve is somewhat well-known; it is called the folium of Descartes.
There might be a sleeker way to do this, but I find that the usual curvature formula for plane curves works nicely enough: to find the curvature of the curve y = f(x) at the point (x0, f(x0)), we have
\(k(x_0)= \frac{|f''(x_0)|}{|1 + f'(x_0)^2|^{3/2}}\)
Even if our curve is not exactly a one-to-one function, locally, we can say it is, and implicit differentiation will save us here. Let’s pretend that y = f(x) for some f around the point (3a/2,3a/2). Then, by differentiating both sides of the equation with respect to x, we get
3x2 + 3y2y′ = 3ay + 3axy′
and isolating y’ yields
\(y' = \frac{3x^2 - 3ay}{3ax - 3y^2}\)
\(= \frac{x^2 - ay}{ax - y^2}\)
Then, to get y′′, we just differentiate y′ with respect to x:
\(y'' = \frac{(ax - y^2)(2x - ay')-(x^2 - ay)(a - 2yy')}{(ax - y^2)^2}\)
\(y'' = \frac{2ax^2y}{(ax - y^2)^2}\)
Substituting \(x = y = \frac{3a}2\) gives us \(y'= -1, y'' = -\frac{32}{3a}\) and so
\(k(\frac{3a}2) = \cfrac{\frac{32}{3a}}{2\sqrt 2}\)
\(= \frac{8\sqrt 2}{3a}\)
The radius of curvature is, \(\frac 1k= \frac{3a}{8\sqrt 2}\).
