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.
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.