With regard to the following image above:
Let there be an ellipse with center $O$, vertices $A, A'$, co-vertex $B$, foci $F, F'$.
Draw $FP$, perpendicular to the major axis (semilatus rectum).
Draw the tangent to the ellipse at $P$, meeting the major axis at $T$
Apparently, $T$, a point on this tangent line, lies at the intersection between the major axis and the directrix. Is there a synthetic geometric proof for this?
I have $\angle TPF = \angle F'PQ$ by the reflective property and thus $\triangle TPF \sim \triangle F'PQ$. Then I'm stuck.
Let $T$ be the point where the tangent at $P$ meets the directrix. We want to show that $T$ also belongs to line $AA'$, i.e. that $MT=PF$, where $M$ is the projection of $P$ onto the directrix (see figure below).
Draw normal $PN$, perpendicular to the tangent and meeting $AA'$ at $N$. From the angle bisector theorem we have: $$ FN:PF=FF':AA'=e=PF:PM, \quad \text{where $e$ is the eccentricity of the ellipse.} $$ But triangle $PFN$ is similar to $PMT$, hence: $$ MT:PM=FN:PF=PF:PM,\quad\text{that is:}\quad MT=PF, $$ Q.E.D.