[vlasev]

Let P be the point (a,b) and let Q be the point (x(t),y(t)) of a curve for some parameter t. The tangent vector to that curve is (x'(t),y'(t)) and the ray PQ is just (a-x(t),b-y(t)). You need to make sure these vectors are the same (up to a constant multiple) in order to have a tangent ray. That is you are led to two equations

cx'(t) = a-x(t) cy'(x) = b-y(t)

You need to solve these two equations for values of c and t. The difficulty of this task depends mostly on the difficulty of the curve itself.

If the curve has equation y = f(x) (or in other words a simple representation x(t) = t and y(t) = f(x(t)) = f(t)), then you have

c = a-x cf'(x) = b-f(x)

Therefore (a-x)f'(x) = b-f(x). If f(x) is a polynomial of degree n, the problem is reduced to root-finding of a polynomial of degree n so you'll have at most n roots to deal with. So in general it won't be easy. This reduces to the code in the post if you use line segments.