TITLE: Non-trivial questions about confocal ellipses
QUESTION [12 upvotes]: This question arises from an attempt to give an elementary solution to an interesting problem.


Main question. Assume that $\Gamma_{AB}$ and $\Gamma_{AC}$ are two ellipses with foci at $A,B$ and $A,C$ respectively, meeting at $P$.
  Let $Q$ be the other point of intersection of $\Gamma_{AB}$ and
  $\Gamma_{AC}$: how to construct $Q$ with straightedge and compass,
  given $A,B,C,P$?

My attempt was to write the equation of both ellipses in polar coordinates and perform some trigonometry, but that does not translate easily into a simple straightedge and compass construction. So I wondered about some hidden projective property of two confocal ellipses, and indeed something striked me.

Secondary question. In the previous configuration, let $U$ be the pole of $PQ$ with respect to $\Gamma_{AB}$, i.e. the intersection of
  the tangents at $P$ and $Q$, and let $V$ be the pole of $PQ$ with
  respect to $\Gamma_{AC}$. Then $A$ lies on $UV$.

I was not able to prove this fact, but I discovered many interesting theorems in this Dolgirev paper, and I believe his techniques solves both the secondary question and the main question. May I ask some help from you, guys?

REPLY [4 votes]: This is not an answer. It's just too long and too illustrated for a comment.
I prefer to do most of my projective geometry over $\mathbb C$, so I'll be viewing the problem from this point of view. What does it mean for a point $A$ to be a focus of a given conic? It means that the lines joining $A$ to the ideal circle points $I=[1:i:0]$ and $J=[i:1:0]$ are tangents to the conic. So when you are talking about a pair of two confocal conics, then a more general description would be speaking about a pair of conics sharing two tangents. The common focus $A$ is the point where these tangents intersect. To make this easier to imagine, for the following picture I drew the points $I$ and $J$ at real positions. The brown lines are common tangents for the red and the green conic.

As Piquito pointed out, two conics will in general have four points of intersection. In the case of confocal ellipses it might well be that two of them have to be complex. But in the illustration above, all four of them are real.
If you pick the correct pair $P,Q$, then the line joining their poles does indeed pass through $A$, as you claimed. The same is true for the opposite pair of intersections, namely $R,S$. Their poles lie on the same line as those for $P,Q$. For other pairs of intersection, however, that is not the case. The other poles (red points) lie on two different lines (orange), four poles on each (not all of which are visible in the section above).
At this point I don't know how this may help you. Perhaps the most valuable property here might be the fact that the line joining the poles of $PQ$ is concurrent with the lines $PS$ and $QR$. Since $R$ and $S$ are complex in your setup, that does not lead to an obvious construction, but it might help with a proof for your second question. The fact that you have to match things correctly may make an algebraic proof somewhat complicated, though.