$A,B,V$ are three distinct, non-collinear points in the plane and we want to find, through straightedge and compass, the focus $F$ of a parabola $\wp$ with vertex at $V$ and going through $A,B$.
Context: it is not difficult to prove that the solution is unique. We need five linear constraints to fix a projective conic: "$\wp$ is a parabola" is equivalent to "the center of $\wp$ lies on the line at infinity" and provides the first one; $A,B,V\in\wp$ provide the second, third and fourth constraints; "$V$ is the vertex of $\wp$" is equivalent to "the normal line at $V$ meets the line at infinity at $\wp$" and provides the fifth one.
Additionally, in some reference system centered at $V$ we have $A=(u,au^2), B=(v,av^2)$, hence if we manage to compute $|a|$ from the squared side lengths of $ABV$, namely
$$u^2+a^2 u^4,\qquad v^2+a^2 v^4,\qquad (u-v)^2+a^2(u-v)^2(u+v)^2$$
we are done. On the other hand, it does not seem so trivial to do that by straightedge and compass.
This page collects a lot of facts on the Euclidean geometry of parabolas, that might be useful in tackling the current problem.
If $AV=4, BV=5$ and $AB=6$ then $a^2$ turns out to be a root of the cubic polynomial $$-28672-1229504 x-8266300 x^2+121550625 x^3$$ which is irreducible over $\mathbb{F}_{11}$, hence irreducible over $\mathbb{Q}$.
It follows that the given problem cannot be solved by straightedge and compass.