When playing around with conics in GeoGebra, I have found out that the following relation seems to hold:
Let parabola $p$ be tangent to sides/extensions of sides $BC,CA,AB$ of triangle $ABC$ at points $P,Q,R$. Call $F$ the focus of this parabola. Let $T$ be intersection point of lines $AP,BQ,CR$ (they are concurrent as a corollary to Brianchon theorem) and let $S$ be the intersection point of circumcircle and Steiner's circumellipse other than $A,B,C$ (a.k.a. Steiner point). Then $F,T,S$ are collinear.
My question is: can anyone provide a proof of this relation? I imagine this would be quite a complex result to show, but maybe it already exists somewhere in literature (in which case I'd be thankful for a reference).
Thanks in advance.
Illustration:
