I found a geometry question posted today very interesting and would like to clear a doubt regarding a parabola if it can be inscribed (as described below).
Setup : Two equilateral triangles $\triangle ABC$, $\triangle CDE$ stand on $\overline{BD}$ as shown. $AD, BE$ intersect at $F$. Show that $FC^2 = AF \cdot EF$
If tangents at two points $A,E$ on the parabola meet at $C$, $F$ being the focus of parabola, then $FC^2=AF \cdot EF$ holds. Now if I try to inscribe a parabola in the region of $\angle ACE$ such that it is tangent to $AC$ at $A$ and $CE$ at $E$, does the condition $FC^2=AF \cdot EF$ uniquely identifies $F$ as focus of such a parabola, or are there more than one parabola with different focus also possible?
I thought that five conditions should uniquely determine any conic. Those known conditions here are points $A,E$, tangents at $A,E$ and the property that $FA, FC, FE$ are in geometric progression. But I am not sure if this direction of last condition is sufficient since perhaps we can find $F'$ somewhere else for which similar triangles $AF'C, CF'E$ can be made, and $F'$ will become the focus of second such parabola.
To be precise, can somebody suggest a proof for the converse of the property that if $FC^2 = AF \cdot FE$, where $AC, EC$ are tangents to the parabola, then $F$ must be the focus of such parabola.
Here is one such parabola with focus $F$ and meeting above conditions.
Bonus for those interested in a fun challenge in this setup, not related to the problem above :
Let $BE$ cut $AC$ in $P$ and $AD$ cut $CE$ in $Q$. Given $AF = 4, FC = 6$, calculate with proof, $BP + DQ$. Answer : $19$
Unfortunately the converse is false: there are infinitely many points $F$ such that $FC^2=AF\cdot FE$, but only one of them is the focus of the parabola tangent to lines $AC$, $EC$ at points $A$, $E$ respectively.
To see why, choose any point $P$ on $AC$ and construct $Q$ on $EC$ such that $$ CQ:QE=AP:PC. $$ We can then construct the circle of Apollonius through $P$ whose points have all the same ratio of distances from $A$ and $C$, and also the circle of Apollonius through $Q$ whose points have all the same ratio of distances from $C$ and $E$. If $F$ is an intersection of those circles, then: $$ AF:FC=AP:PC=CQ:QE=FC:FE $$ and consequently $FC^2=AF\cdot FE$.
Varying the position of $P$ we can thus construct many points having that property. But only one of them is the focus $G$ of the parabola tangent to lines $AC$, $EC$ at points $A$, $E$, which is uniquely determined.
EDIT.
We can construct the focus $G$ of the parabola as follows. Draw a line $r$ through $C$ and through the midpoint of $AE$: this line is parallel to the axis by a property of the parabola.
Draw two lines $s$, $t$ parallel to $r$ and passing through $A$, $E$ respectively. By the reflective property of the parabola, the reflection of $s$ about the tangent at $A$ passes through the focus, and the same is true for the reflection of $t$ about the tangent at $E$. Hence focus $G$ is the intersection of those reflected lines.