Let the vertices of a triangle $T$ be $(A,B,C)$, and $(a,b,c)$ the vertices of its Morley triangle $M$. Designate vertex $C$ as the apex of $T$. Now move apex $C$ parallel to $AB$, all the while tracking the Morley triangles $M(C)$ of the modified $T(C)$, where the notation indicates that both $T$ and $M$ are functions of the position of $C$.
As $C$ moves far to the left of $A$, $M(C)$ approaches $A$, and similarly $M(C)$ approaches $B$ as $C$ moves far to the right. Indeed all three vertex curves approach $A$ and $B$ My question is:
Q. What are the curves that the three vertices of $M(C)$ follow?
They look something like ellipses, but I do not believe they are ellipses, for they intersect too frequently. E.g., see $b$- and $c$-paths below:
The curves are definitely not ellipses. I'll just discuss the bottom corner, $a$ (which I'll call $F$).
Coordinatize, with $A = (-p,0)$, $B = (p, 0 )$, $C = (c, h )$. Let the "lower" trisectors from $A$ and $B$ meet at the "bottom" Morley vertex, $F = (x,y)$. Then $$\tan \angle CAB = \frac{h}{p+c} \qquad \tan\angle CBA = \frac{h}{p-c}$$ $$\tan \angle FAB = \frac{y}{p+x} \qquad \tan\angle FBA = \frac{y}{p-x}$$
Leveraging the triple-angle formula for tangent, $$\tan 3\theta = \frac{\tan\theta\;(\;3 - \tan^2\theta\;)}{1 - 3 \tan^2\theta}$$ we obtain $$\begin{align} \angle CAB = 3\angle FAB:\quad c y (\;3 ( x+p )^2 - y^2\;) &= h ( x + p )^3 - 3 p y ( x + p )^2 - 3 h y^2 (x+p) + p y^3 \\ \angle CBA = 3\angle FBA:\quad c y (\;3 (x-p)^2 - y^2\;) &= h (x-p)^3 + 3 p y ( x-p )^2 - 3 h y^2 (x-p) - p y^3 \end{align}$$ Eliminating parameter $c$, and doing a bit of clean-up, we get the relation defining (a superset of) the locus of $F$ as triangle vertex $C$ moves at distance $h$ from $\overleftrightarrow{AB}$: $$3 h ( x^2 + y^2 - p^2 )^2 - y ( 3 x^2 - y^2 - 3 p^2 )^2 - 4 p^2 y^2 (h - 3 y) = 0 \qquad (\star)$$
Here's a picture of the solution set of $(\star)$, with $p=h=1$. The Morley locus is highlighted in red.