TITLE: Partition of plane into parabolas
QUESTION [10 upvotes]: The plane is partitioned into parabolas (each point belongs to exactly one parabola). Does
it follow that their axes have the same direction?

REPLY [3 votes]: Lemma 1: If parabola A is inside parabola B, then A and B must have parallel axes.
("Inside B" means on the side that B curves towards, in other words, the convex hull of B is B and its "interior".)
Zooming out, a parabola looks more and more like a ray.  For example, if we look at $y=x^2$, then zooming out by a factor of 100 (i.e. $y_{zoom}=y/100$ and $x_{zoom}=x/100$) turns it into $y_{zoom}=100x_{zoom}^2$.
If A's axis is not parallel to B's, then if we zoom out sufficiently, we will see A leaving the interior of B.
But since A and B share no points, this cannot happen, so A's axis must in fact be parallel to B's.
(From the projective point of view, having non-parallel axes means touching different points on the line at infinity, which clearly can't happen if one is inside the other.)
Parabolic Regions
Given a parabola $A$ in the partitioning of the plane, let $R(A)$ be the union of all parabolae which contain $A$ or are contained by $A$.
The notion of containment provides an ordering on these parabolae.
If the region $R(A)$ is the entire plane, then by lemma 1 all the axes are parallel.
If the region $R(A)$ is not the entire plane, then (considering the supremum under the ordering) it must be convex with a parabolic boundary (either open or closed).
[[EDIT: $R(A)$ could also be an open half-plane. If two regions are open half-planes, then no parabolae can fit in the gap between them, so at most one region can be an open half-plane. The following paragraph rules out even this possibility (since although the open half-plane region is not strictly convex, the other regions still are).]]
If it is open (not including its parabolic boundary), then every point on the boundary must be a member of another region.  Since the regions are strictly convex, no two points on the boundary can be members of the same other region, and so an uncountable number of other regions are needed.  However, since each region includes a rational point in its interior, there can only be a countable number of regions.  Therefore, there cannot be an open region.
A Reduced Question
So the original question reduces to the question of whether it is possible to partition the plane into convex closed parabolic regions.
Such a partitioning is not possible. (We will follow Cantor here rather than Sierpiński).  Since we can enumerate the rational points, this yields an enumeration of the parabolic regions (ordered by earliest interior rational point).  This lets us define a sequence of shrinking open intervals on an arbitrary line: Starting with any interval I$_0$ that spans multiple parabolic regions, we define the next open interval I$_{n+1}$ as the interval between the first two parabolic regions (according to the enumeration) to intersect the interval I$_n$.
These intervals shrink towards either a limit point or a limit interval, and either way provides a point contained in all intervals.  This point cannot be in any parabolic region -- the parabolic region would have appeared in the enumeration and so would have been used at some point to reduce the interval.
So the answer to the reduced question is no, it is not possible to partition the plane into convex closed parabolic regions.
Therefore the answer to the original question is yes, if the plane is partitioned into parabolas, then it follows that their axes must all have the same direction.