TITLE: Parabolas through three points
QUESTION [15 upvotes]: We can draw an infinite number of parabolas that pass through three given points $A$, $B$, $C$ (in that order). For each such parabola, we take the tangent lines at $A$ and $C$, and intersect them to get a point, $P$, which is called the "apex" of the parabola segment, in my business. What is the locus of this apex point $P$? I think it's a hyperbola that has the lines $AB$ and $BC$ as asymptotes.
Here's a picture. The pink curve is the locus that interests me.

I believe the point $P$ is called the "polar point" of the chord $AC$, sometimes.  
The answers to this question have some basic computations that might be useful.
The locus of the parabola vertex seems to be much more complicated. It is studied in this question. Given the complexity of the vertex locus, I think it's remarkable that the apex locus is so simple.
A parabola is, of course, a quadratic Bezier curve, and the apex point is its "middle" control point (if this helps).
Can we generalize to rational quadratic Bezier curves (i.e. to conic section curves that are not necessarily parabolas)?
Edit:
The generalization to rational quadratic Bezier curves probably doesn't make sense. You can (often) pick any two additional points, say $D$ and $E$, and you'll be able to draw a conic through all five points $A, B, C, D, E$. So, the locus of the apex point is some region of the plane, not a curve.

REPLY [2 votes]: Here's a fairly quick algebraic approach to @MvG's "one step further" simplification. (This was actually my first approach.) We leverage the fact that tangency, parabola-ness, and hyperbola-ness are preserved under affine transformations, and we apply such a transform to move the defining points to coordinates
$$A(1,0) \qquad B(0,0) \qquad C(0,1)$$
Now, instead of using the general equation of the conic to describe the locus as @MvG did, I'll use (slightly different version of) it to describe the parabola.
$$a x^2 + b y^2 + 2 h x y + 2 d x + 2 e y + c = 0 \qquad (1)$$
We can make several observations about the coefficients of $(1)$.

Since $A$, $B$, $C$ are on the parabola, we substitute their coordinates into $(2)$ to get these relations
$$\begin{align}a + 2 d + c &= 0\\
c &= 0\\
b + 2 e + c &= 0
\end{align}$$
whereupon $(1)$ becomes
$$a x^2 + b y^2 + 2 h x y - a x - b y = 0 \qquad (2)$$
Next, we reference known "pole and polar" lore. Point $P$ is a pole of its parabola; and $AC$ is the corresponding polar (line). As the polar, line $AC$ has an equation derived from $(2)$ to be
$$p x + q y + r = 0$$
where
$$
p = a u + h v - \frac{1}{2}a \qquad
q = h u + b v - \frac{1}{2}b \qquad
r = - \frac{1}{2} a u - \frac{1}{2}b v$$
We know that $A(1,0)$ and $B(0,1)$ lie on this line, so
$$p + r = 0 = q + r$$
Solving the system for $u$ and $v$ gives
$$u = \frac{b}{2h} \qquad v = \frac{a}{2h} \qquad \implies \qquad u v = \frac{a b}{4h^2}$$
Finally, simply recall (or look-up) that, for $(1)$ to represent a parabola, we must have $h^2 = a b$. Therefore, $$u v = \frac{1}{4}$$ The locus of $P(u,v)$ is indeed a hyperbola with $AB$ and $BC$ (aka, the coordinate axes in our affinely-transformed case) as its asymptotes.