An ellipse tangent to a kite

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Let \(F_1\) and \(F_2\) be the two focal points. Recall the optical property of the ellipse: for any point \(P\) on the ellipse, the lines \(F_1P\) and \(F_2P\) makes the same angle with the tangent at \(P\).
Therefore, in the diagram above, \(\angle F_1 P_1 Q = 180 - \angle F_2P_1 Q\). Since the figure is symmetric after reflecting across the minor axis, then this equation becomes \(\angle F_1 P_1 Q = 180 - \angle F_1 P_2 Q\). This is precisely the condition for the quadrilateral \(F_1 P_1 Q P_2\) to be cyclic, so therefore \(F_1\) lies on the unique circle through \(P_1\), \(P_2\) and \(Q\).
The same idea shows that \(F_2\) also lies on this circle. Therefore \(F_1\), \(F_2\), \(P_1\), \(P_2\) and \(Q\) are concyclic.
If you know the line through the major axis (which we constructed in class) then you can construct the focal points as the intersection points of this line with the circle through \(P_1\), \(P_2\) and \(Q\).