Classification of quadratic rational Bézier curves

Question

My teacher months ago gave me a few hints on a method that can classify quadratic rational Bézier curves as different conic sections (arcs of those).

1. As I recall, it starts with such a curve given three points.
2. Then via a reparametrization it makes the first and last coefficients the same (I remember it being a mobius transformation), in this way one can simplify them leaving only the middle one $\beta$.
3. After that, it uses some affine geometry, defines an affine coordinate system using the three points.
4. In the end finds out the standard cartesian form of a conic and studies the coefficients as functions of the previous middle coefficient $\beta$.

Can someone help me reconstruct the whole method? Thanks

PS: I built myself a similar method. I start by computing the parametric coordinates of a Bézier curve of second degree(not a rational one). Then I regroup the powers of the parameter and I apply an "implicitization", but this process is neither elegant or quick and not the one the teacher gave me.

Answer 1

What your teacher was referring to is the fact that:

$$x=\dfrac{a t^2+bt+c}{gt^2+ht+k}, y=\dfrac{d t^2+et+f}{gt^2+ht+k}, \ \ \ 0 \le t \le 1 \tag{1}$$

(with real coefficients $a,b,c,d,e,f,g,h,k$)

is in general a conic curve under the form of a rational quadratic Bezier curve.

This property comes from the vast domain of projective geometry.

The classification is easy: either the denominator

• can be zero twice (which happens when its discriminant $\Delta=h^2-4gk > 0$) it means that there are two points at infinity: you have a hyperbola (a hyperbola has 4 points at infinity, but in projective geometry opposite points are considered as identical).

• can be zero once ($\Delta=0$): parabola.

• can never be zero ($\Delta<0$): ellipse.

In order to retrieve the driving points $A,B,C$, just decompose the numerators onto the basis $(1-t)^2, 2t(1-t), t^2$, which means finding $x_A, x_B,...$ such that:

$$\begin{cases}a t^2+bt+c&=&(1-t)^2 x_A + 2t(1-t)x_B+t^2x_C\\d t^2+et+f&=&(1-t)^2 y_A + 2t(1-t)y_B+t^2y_C\end{cases}$$

Remark: Any bijective change of parametrization $T=f(t)$ in (1) such that

$$f(0)=0 \ \ \text{and} \ \ f(1)=1\tag{2}$$

is possible. Taking a "Möbius" transformation

$$T=\frac{t}{rt+(1-r)}$$

complying with (2) preserves the form of equations (1) and allows, for an adequate choice of parameter $r$, to obtain (slightly) simpler expressions.

Recall: Equations (1) can be written under the form:

$$\underbrace{\begin{bmatrix} X\\ Y\\ Z\end{bmatrix}}_{C'}=\underbrace{\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & k \end{bmatrix}}_H \underbrace{\begin{bmatrix} t^2\\ t\\ 1 \end{bmatrix}}_C \tag{2}$$

followed by :

$$x=X/Z \ \ \text{and} \ \ y=Y/Z \tag{3}$$

Equation (2) expresses that one takes the image of "standard" parabola with parametric description given by vector $C$ (with so-called "homogeneous coordinates" : please note the third coordinate equal to $1$) to which an "homography matrix" $H$ is applied. Equations (3) are the classical homogenizing ratios by the third coordinate. See Fig. below.

Knowing that the image by a homography (either under the form (1) or (2)+(3)) of a conic curve is a conic curve.

See as well here.