Two parabolas $P_1$ and $P_2$ with the same focus $F$ are given so that the vertex tangent of $P_1$ is also the directrix of $P_2$. Let $t_1$ and $t_2$ be parallel lines such that $t_1$ touches parabola $P_1$ in $D_1$, and $t_2$ parabola $P_2$ in $D_2$. Prove that $F$, $D_1$ and $D_2 $ are collinear.
Let $D_1$ be the touching point of parabola $P_1$ and tangent $t_1$. Let $H$ be the intersection point of $t_1$ and $d_1$ where $d_1$ is directrix of $P_1$. Now if we find axisymmetric image $F'$ of $F$ in relation to $t_1$ and mark the intersection point of $FF'$ and $t_1$ as $G$ , we can see that the triangle $GHF'$ is similar to the triangle $GF'D_1$, i.e. the triangle $GFD_1$. Since $G$ also lies on the directrix $d_2$ of the parabola $P_2$, it follows that $G=t_1 \cap d_2$. Now we construct the tangent $t_2$ of the parabola $P_2$ parallel to $t_1$. We look at the intersection point of a perpendicular line from $F$ to $t_1$ and directrix $d_2$. But since $G=FF' \cap t_1\cap d_2$ , it follows that $G$ is our desired point. The bisector of length $FG$ is $t_2$. Let $I$ be the intersection point of $t_2$ and $d_2$, and $J$ the intersection point of $FG$ and $t_2$. Now we observe that the triangle $JIG$ is similar to the triangle $JFD_2$. But since $JIG$ is similar to the triangle $GHF'$, and $GHF'$ is similar to $GFD_1$, we arrive at the fact that $JFD_2$ is similar to the triangle $GFD_1$, which means that $\angle JFD_2$=$\angle GFD_1$, hence $F$, $D_1$ and $D_2$ are collinear.
I think my proof is all over the place, it's too long and it's missing some crucial steps. I don't even know if it's valid. I would appreciate your corrections and suggestions on how to deal with these kinds of problems in general.
