Possibly rotated parabola from three points

I am pretty sure there is no simple solution for this problem.

You can assume the origin is at the vertex via the transformation $x\mapsto x-v_1$, $y\mapsto y-v_2$.

Via a rotation you can assume that the equation is $\bar x^2=a \bar y$. The rotation is given by an angle $\alpha$, or equivalently, by $s=sin(\alpha)$ and $c=cos(\alpha)$, with $c^2+s^2=1$.

Then $$ \bar x=c x+ s y,\quad \bar y = -s x+c y,\quad \text{(and so $x=c\bar x-s\bar y,\quad y=s\bar x +c\bar y$)} $$ which gives the equation of the parabola in the original variables as $$ (cx+sy)^2=a(-sx+cy). $$ Evaluating this equation at $P=(p_1,p_2)$ gives $$ a=\frac{(c p_1+s p_2 )^2}{c p_2-s p_1}. $$ If you insert this in the equality $$ (cq_1+sq_2)^2=a(-sq_1+cq_2),\qquad\text{(parabola equation at $Q=(q_1,q_2)$)} $$ and multiply by $(-sp_1+cp_2)$, you obtain the third degree equation $$ Ac^3+Bc^2s+Cc s^2+Ds^3=0, $$ with $$ A=p_2 q_1^2-p_1^2 q_2,\quad B=(p_1-q_1)(p_1 q_1-2 p_2 q_2), $$ $$ C=(p_2-q_2)(2 p_1 q_1- p_2 q_2),\quad\text{and}\quad D=p_2^2 q_1-p_1 q_2^2. $$ It is easy to see that $AD\ne 0$ if $P\ne V$, $Q\ne V$ and the three points are not aligned, hence we have a third degree equation either for $\frac cs$ or $\frac sc$ (maybe for both).

Assume you solve this equation for $\frac cs$ and obtain $\frac cs=K$. Then $$ s=\frac{1}{\sqrt{K^2+1}}\quad\text{and}\quad c=\frac{K}{\sqrt{K^2+1}}, $$ which gives you the complete solution (i.e., the equation) of the parabola, since you have $a$ in terms of $c$ and $s$.

This is NOT a construction you can carry out with ruler and compass in general, since the values have minimal polynomials up to degree six.

For example, if $P=(1,2)$, $Q=(1,3)$, then $a\sim 0.0132968$ is a zero of the polynomial $$ P_a(x)=50 x^6-35 x^4+11 x^2-1, $$ $s=sin(\alpha)\sim -0.377016\ $ is a zero of the polynomial $$ P_s(x)=50 x^6-165 x^4+5656 x^2-1, $$ and $c=cos(\alpha)\sim 0.926207\ $ is a zero of the polynomial $$ P_c(x)=50 x^6-115 x^4+91 x^2-25. $$

So, in general you have to solve a third degree equation, which I think is not simple. You can always choose $a$ to be greater than zero, and you must choose solutions of $c,s$ such that $\bar P_x \bar Q_x=(cp_1+sp_2)(cq_1+sq_2)<0$, in order that $V$ is between $P$ and $Q$.

Maybe one can use the Origami constructive geometry to do this in a simpler way.

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.