On implicit equations of rational quadratic Bézier curves

Question

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.

Answer 1

Something is fishy here. You can easily have two different conic sections which have the same tangents at two points, but are not the same. Here is an example:

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.