Solved it myself, as below. I don't feel that this solution is "right" somehow, because it treats the interior points $\mathbf B$ and $\mathbf C$ very asymmetrically, so improvements are invited. But, anyway ...
Let's use the notation from the last few paragraphs of the question.
It's clear that $\mathbf P(0) = \mathbf A$, and $\mathbf P(1) = \mathbf D$, so these two points are interpolated, already, and we only have to worry about the other two points, $\mathbf C$ and $\mathbf D$.
First, we find numbers $h$ and $k$ such that
$$\mathbf C = \mathbf A + h(\mathbf B - \mathbf A) + k(\mathbf D - \mathbf A)
= (1-h-k)\mathbf A + h\mathbf B + k\mathbf D$$
This is possible provided that $\mathbf A$, $\mathbf B$, $\mathbf D$ are not collinear. Then, since $\mathbf P(v) = \mathbf C$, we have
$$(1-h-k)\mathbf A + h\mathbf B + k\mathbf D =
(1-v)^2 \mathbf A + 2v(1-v) \mathbf P + v^2 \mathbf D $$
But, since $\mathbf P(u) = \mathbf B$, we know that
$$ \mathbf B = (1-u)^2 \mathbf A + 2u(1-u) \mathbf P + u^2 \mathbf D$$
Substituting for $\mathbf B$ on the left-hand side, and equating coefficients of $\mathbf A$, $\mathbf P$, $\mathbf D$ gives
$$(1-v)^2 = 1 - h - k +h(1-u)^2 $$
$$2v(1-v) = 2hu(1-u)$$
$$v^2 = hu^2 + k $$
We can easily eliminate $v$ from these last three equations using the fact that $[2v(1-v)]^2 = 4[v^2][(1-v)^2]$. We get:
$$[2hu(1-u)]^2 = 4[hu^2 + k][1 - h - k +h(1-u)^2]$$
After a little algebra, this reduces to:
$$h(1-h)u^2 - 2hku + k(1-k) = 0$$
So, we solve this quadratic for $u$, and then get the unknown interior control point $\mathbf P$ from
$$ \mathbf P = \frac{\mathbf B - (1-u)^2 \mathbf A -u^2 \mathbf D}{2u(1-u)}$$
The number of solutions depends on the number of real solutions of the quadratic. Quite often, you can draw two different quadratic Bezier curves through the four given points, as explained in the paper cited in the question.
Given a point $(a,b)$ and a horizontal line $y=k$, in $\mathbb{R^2}$.
Let the locus of the points which are equally far away from the point and the line be denoted by $(x,y)$.
Then distance between $(x,y)$ and the line is just $|y-k|$
The distance between the point $(x,y)$ and $(a,b)$ is $\sqrt{(x-a)^2 + (y-b)^2}$
Equating the two:
$|y-k|=\sqrt{(x-a)^2 + (y-b)^2}$
$(y-k)^2=(x-a)^2+(y-b)^2$
Expanding and rearranging:
$y=\frac{x^2-2ax+(a^2+b^2-k^2)}{2b-2k}$
Given any quadratic function, you can find the unique value for $a,b,k$ (thus the diretrix and focus).
Best Answer
The general parabola $y = f(x)$ with a given root $p$ (i.e., given $x$-coordinate of the endpoint) has the form $$y = f(x) = - A (x - p) (q - x),$$ where $A$ and $q$ are arbitrary constants. The condition that the parabola passes through $(a, b)$ is that $f(a) = b$, so that $$\boxed{A(a - p)(a - q) = b} .$$ By symmetry the vertex of the parabola has $x$-coordinate $\frac{p + q}{2}$, so the requirement that the parabola has (for $A > 0$) maximum height (i.e., $y$-coordinate) $h$ there is $f\left(\frac{p + q}{2}\right) = h$, or $$\boxed{A (p - q)^2 = -4 h} .$$