TITLE: A circle and an ellipse
QUESTION [9 upvotes]: Consider an ellipse and a circle inside it. They have the same center. Let $P$ be a moving point on the ellipse. Through $P$, draw the two tangents to the circle, which meet the ellipse at $A,B$. Again, through $A,B$ respectively, draw the tangents to the circle, which meets each other at $Q$. What's the locus of $Q$?
By software graphing, it seems to be an ellipse too. But how to prove this without much computation? It's better to explain by projective geometry.

REPLY [3 votes]: The problem can be generalized so that both curves (the circle and outer ellipse  in the diagram) are conics. Then the resulting locus is also a conic. Although the subject matter and concepts are very much that of projective geometry, the methods of proof tend to be more analytic than synthetic.
As described in Andrea Del Centina's Poncelet’s porism: a long story of renewed discoveries (also on JSTOR) the problem was studied in the 19th century, and is related to Poncelet's Theorem.  
Look for  Fig. 21 in Del Centina's paper (pg 68) which illustrates the approach in George Salmon's On the problem of the In-and-circumscribed Triangle. 

Salmon derives an equation for the locus as a weighted sum of the original conics ($U$ and $V$) and a third conic $F$. ($F$  has the remarkable property that the tangents from any point on $F$ to $U$ and $V$ form a harmonic pencil.)
Further on, look for Fig. 31, which illustrates a theorem by Darboux.

A tangential pencil is the set of all conics tangent to a four given lines (some of which may be complex), and is dual to an ordinary conic pencil: the set of all conics containing four given points (some of which may be complex). 
If $K_3=K_2=K_1$,  the quadrilateral circumscribes $K$ (the inner circle/conic in the question) and three of its vertices lie on $K_1 $ (the outer ellipse/conic), as in the original question (see figure below).  Then Darboux's theorem says that the locus of the fourth vertex $Q$ is a conic in the tangential pencil $\mathcal T$ containing $K$ and $K_1$.

To compute the members of $\mathcal T$, let $K'$ and $K'_1$ be the dual conics of $K$ and $K_1$.  Then $\mathcal T$ is the set of duals of members of the pencil containing $K'$ and $K'_1$.
Finally, the paper points out that Darboux observed that his theorem was the dual of Poncelet's General Theorem.  In the comments, @JeanMarie has pointed out that this question has some similarity with another question math.stackexchange.com/q/3509582.  And indeed that one is the dual of this one and is a case of Poncelet's General Theorem.