I'm given an ellipse with focal points $F_1$ and $F_2$, and two points $P$ and $Q$ on the ellipse such that $F_1$ lies on $PQ$. Also, the tangents at $P$ and $Q$ intersect at a point $R$.
Show that:
(a) $R$ is the excentre of the triangle $\Delta PQF_2$.
(b) The excircle is tangent to $PQ$ at $F_1$.
I know that to show that a point is an excentre, I'd need to show that the point is the intersection of three angle bisectors. However, I have no idea how to show that I have the three angle bisectors. I'd really appreciate any help given.
Let us denote a couple of points on your picture. Take $M$ to be on $RP$ such that $R-P-M$ and $N$ on $F_2P$ such that $F_2-P-N$.
By optical property of ellipse $\angle F_1PR= \angle F_2PM$, so since $\angle F_2PM=\angle RPN$ we have $\angle F_1PR= \angle RPN$, i.e. $RP$ is the bisector of $\angle F_1PN$, i.e. of the external angle at $P$ in $\triangle PQF_2$. Similarly, $RQ$ is the bisector of the external angle at $Q$ in $\triangle PQF_2$. Therefore, $R$ is the excenter of $\triangle PQF_2$.
For the second part recall the following: In $\triangle ABC$, the excircle tangent to the edge $BC$, touches $BC$ in the point $K$ such that $BK=s-BA$, where $s$ is the semiperimeter.
In $\triangle PQF_2$, by the definitional property of the ellipse, $PF_1+PF_2= QF_1+QF_2$, so the semiperimeter is actually $s=PF_1+PF_2$, now $PF_1=s-PF_2$, so by the previous observation $F_1$ is the point where the excircle is tangent to the edge $PQ$.