From

Bezier curve - Wikipedia, the free encyclopedia, the parametric equation for a quadratic Bezier curve is

http://upload.wikimedia.org/math/b/0...30f26f60b0.png
From

Parabola - Wikipedia, the free encyclopedia, the general equation for a parabola is

http://upload.wikimedia.org/math/4/7...baa30a1394.png
and there are further conditions, including $\displaystyle B^2 = 4AC$.

I don't know for a fact that a quadratic Bezier curve determines a parabola, but I'll assume it does. Given the condition on $\displaystyle B$, it cannot assumed that $\displaystyle B = 0$ as you indicate.

Given the 3 points defining the Bezier curve, you want to determine the 6 coefficients of the parabola. Under the invertibility requirement below, you may set the coefficient $\displaystyle F$ arbitrarily to a non-zero number, say $\displaystyle F = -1.$ Then you need 5 linear equations to determine the other 5 unknown coefficients.

Let $\displaystyle (x_i,y_i) = \bold{B}(i/4),\ i = 0,\ldots, 4.$

Then the required 5 linear equations are

$\displaystyle \begin{bmatrix}

x_0^2 & x_0 y_0 & y_0^2 & x_0 & y_0 \\

x_1^2 & x_1 y_1 & y_1^2 & x_1 & y_1 \\

x_2^2 & x_2 y_2 & y_2^2 & x_2 & y_2 \\

x_3^2 & x_3 y_3 & y_3^2 & x_3 & y_3 \\

x_4^2 & x_4 y_4 & y_4^2 & x_4 & y_4 \end{bmatrix}

\begin{bmatrix} A \\ B \\ C \\ D \\ E \end{bmatrix} =

\begin{bmatrix} -F \\ -F \\ -F \\ -F \\ -F \end{bmatrix}.

$

To solve these equations requires that the matrix be invertible. That is the invertibility requirement that allows you to set $\displaystyle F$ to an arbitrary non-zero number.

There is a question whether the condition $\displaystyle B^2 = 4AC$ will be satisfied. I think it must if the Bezier curve does indeed determine a parabola, which as I said I'm assuming is true.

Once you have the coefficients, then which side of the Bezier curve a point $\displaystyle (x,y)$ is on is determined by how the sign of $\displaystyle G(x,y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F$ compares to the sign of $\displaystyle G(\bar{x},\bar{y})$ where $\displaystyle (\bar{x},\bar{y}) = \bold{P_1},$ the middle control point for the Bezier curve.