TITLE: Possibly rotated parabola from three points
QUESTION [7 upvotes]: I'm looking for a possibly rotated parabola in the plane, i.e., the solution to a quadric like
$$
Ax^2 + 2Bxy + Cy^2 + Dx + Ey + F = 0
$$
where exactly one of the eigenvalues of 
$$
\begin{bmatrix}
A & B \\ B & C
\end{bmatrix}
$$
is zero, and where the solution set is nonempty. 
I know the $xy$ coordinates of the vertex $V$ of the parabola (i.e., the point that lies on the axis of reflectional symmetry) and of two points $P$ and $Q$ of the parabola, one on each side of the vertex, so that going from $P$ to $V$ to $Q$ along the parabola involves no backtracking. 
(The corresponding problem where $P$ and $Q$ might both be on the same side has multiple solutions for some configurations, but that's why I've included the "between-ness condition".) 
I'd like to know either the coefficients $A, \ldots, F$, or the focus and directrix, or any other unambiguous description of the parabola that I can convert to one of these. 
I keep feeling as if something from projective geometry involving perspectivities and projectivities and (2,2)-correspondences ought to make this simple, but I can't see it.

REPLY [2 votes]: John, I got qualitatively the same result, by what I think is a somewhat different method. I started with $P=(a,b,1)$, $V=(0,0,1)$, $Q=(c,d,1)$ and an arbitrary point $I=(t,1,0)$ on the line at infinity, and drew the conic passing through these four, but tangent to the line at infinity at $I$.
For a rotation about $V$, the indeterminate $t$ is the tangent of the angle necessary to rotate through to bring $I$ to $I'=(0,1,0)$; the corresponding sine and cosine are $t\big/\sqrt{1+t^2}$ and $1\big/\sqrt{1+t^2}$, respectively. The transformed points $P'$ and $Q'$ I’ll not write down here, but the conic passing through $P'$, $V$, $Q'$, and $I'$ will have equation $y=Ax^2+Bx$, no constant ’cause it passes through the origin. And $V$ is the vertex of this parabola if and only if $B=0$.
Plugging the coordinates of $P'$ and $Q'$ into this, you get
$$
B=\frac{\frac{(a-bt)^2}{1+t^2}\frac{ct+d}{\sqrt{1+t^2}}
-\frac{(c-dt)^2}{1+t^2}\frac{at+b}{\sqrt{1+t^2}}}\Delta\,,
$$
of which we only want a condition for zeroness. So the equation for $t$ seems to be the cubic $(a-bt)^2(ct+d)-(c-dt)^2(at+b)=0$.
I tried $P=(-1,1)$, $Q=(5,2)$, and got the equation $-23 + 54t - 12t^2 + 9t^3=0 $, assuming I typed everything in correctly; and of course my hand computations above would need to be checked as well. But I think the principle is valid. For my example, the derivative has negative discriminant, so the polynomial is increasing. At least in this case there is only the one real root.