# On implicit equations of rational quadratic Bézier curves

Rational quadratic Bézier curve with control points $$\boldsymbol{B}_0 = [x_0 : y_0: w_0], \boldsymbol{B}_1 = [x_1 : y_1: w_1], \boldsymbol{B}_2 = [x_2 : y_2: w_2]$$ in homogeneous coordinates of $$\mathbb{R}P^2$$ refer to projective curves of the form $$\boldsymbol{B}(t) = [x(t) : y(t) : w(t)],$$ where $$x(t) = x_0(1-t)^2 + 2x_1 (1-t)t + x_2t^2, \\ y(t) = y_0(1-t)^2 + 2y_1 (1-t)t + y_2t^2, \\ w(t) = w_0(1-t)^2 + 2w_1 (1-t)t + w_2t^2.$$

In Loop-Blinn 2005, section 3.2, it is mentioned that any rational Bézier curve have implicit equation $$c(x,y,w) = k^2 - lm,$$ where $$l$$, and $$m$$ are the homogeneous equations of any two lines tangent to the curve, and $$k$$ is the line connecting these points of tangency. With appropriate choice of functionals $$k,l,$$ and $$m$$, any conic section can be represented in this way.

Problem: How is the appropriate functionals $$k,l,m$$ chosen? Why is $$c(x,y,w) = k^2 - lm$$ a valid implicit equation of the curve?

Example: if we consider the unit circle in $$\mathbb{R^2}$$: $$c(x,y) = x^2 + y^2 - 1 = 0$$, (recall that unit circle is a rational Bézier curve because it admits a parametrization $$\boldsymbol{C}(t) = [1-t^2: 2t: 1+t^2]$$) with points of tangency $$\boldsymbol{P}_1 = (1,0), \boldsymbol{P}_2 = (0,1),$$ then we can set $$l(x,y) = x-1, m(x,y) = y-1, k(x,y) = x + y - 1,$$ so that the line $$l=0$$ is tangent to the circle at $$\boldsymbol{P}_1$$, $$m=0$$ is tangent to the circle at $$\boldsymbol{P}_2$$, and $$k=0$$ is the line connecting $$\boldsymbol{P}_1$$ and $$\boldsymbol{P}_2$$. Homogenizing $$l,m,k$$ with appropriate choice, we get $$l(x,y,w) = x-w, m(x,y,w) = 2(y-w), k(x,y,w) = x+y-w.$$ Then $$k^2-lm = (x+y-w)^2 - 2(x-w)(y-w)\\ = (x^2 + y^2 + w^2 + 2xy -2xw -2yw) - (2xy-2wy-2xw+2w^2)\\ = x^2+y^2-w^2,$$ which is indeed the homogenized equation of the unit circle.

Observations: Plugging in $$\boldsymbol{P}_1, \boldsymbol{P}_2$$ to $$k^2 - lm$$, one get the output $$0$$, so $$\boldsymbol{P}_1, \boldsymbol{P}_2$$ is on the curve $$k^2-lm=0$$. One can also observe that $$k^2-lm=0$$ has tangent lines $$l=0$$ and $$m=0$$ at $$\boldsymbol{P}_1, \boldsymbol{P}_2$$ by calculating the derivative.

• Rational quadratic curves are just conic sections. Commented Jul 2, 2023 at 8:12
• @MarianoSuárez-Álvarez Sorry, I might interpret the paper wrong. In the context of this paper, "quadratic curve" refers to "quadratic Bezier curve", and "rational quadratic curve" refers to "rational quadratic Bezier curve". I will change the description in the question. Commented Jul 3, 2023 at 13:07
• In this way, "quadratic curve" and "rational quadratic curve" are distinct because quadratic Bezier curve can't represent unit circle but rational quadratic Bezier curve can. Commented Jul 3, 2023 at 13:32

We have the circle $$C = \{x^2 + y^2 = 1\}$$, an ellipse $$E = \{ x^2 + 4y^2 = 4\}$$, and a hyperbola $$H = \{ (x+y)(x-y) = 1\}$$. They all touch in the two points $$P = (0,1)$$ and $$Q=(0,-1)$$, and have tangents $$L = \{y = 1\}$$ and $$M = \{y = -1\}$$ there respectively.
Here is another way to see the same problem: If $$l$$ and $$m$$ are linear polynomials which describe the tangent lines at points $$P$$ and $$Q$$ respectively, then so are $$\alpha l$$ and $$m$$ for any $$\alpha \in \mathbb R \setminus \{0\}$$. However, the polynomials $$c = k^2 - lm$$ and $$\tilde c = k^2 - \alpha lm$$ are implicit equations of very different curves!
In the example above, if you choose $$k=x$$, $$l = 1-y$$ and $$m = y+1$$, you obtain the circle with $$\alpha = 1$$, the hyperbola with $$\alpha = -1$$ and the ellipse with $$\alpha = 2$$. Check this!
My best guess is that the authors mean that for some appropriate choice of $$k,l$$ and $$m$$, $$k^2 = lm$$ describes the curve.
• Thank you, this is a good example. Now that I think about it, "$\boldsymbol{P}_1,\boldsymbol{P}_2$ are on the curve, with tangent lines $l, m$" only gives $4$ equations, but a quadratic curve $ax^2+by^2+cw^2+dxy+eyw+fxw=0$ generally have $5$ parameters $b,c,d,e,f$ (because scale doesn't matter, we can throw away a parameter e.g. $a$). So there is $1$ degree of freedom for the quadratic curve to vary. Commented Jul 5, 2023 at 13:29