Finding the curvature of a 2nd degree equation

I have a 2nd degree equation that represents the trayectory of a moving object (in x and y). What i want to find is the curvature ( the reciprocal of the radius of curvature) of said trayectory. I was thinking about finding 3 points in the curve and finding the osculating circle, but I wasnt sure if there was a faster way to do it in mathlab.

Answer 1

[Guillaume]

Not sure what a second degree equation is. Is it a quadratic curve?

Anyway, the wikipedia article on curvature gives you the equations.

If your curve is indeed a quadratic curve, then you can even calculate the curvature analytically. Otherwise, assuming your function is of the form y = f(x),

K = gradient(gradient(y, x), x) ./ (1 + gradient(y, x).^2).^(3/2)