The locus of a triangle's orthocenter - synthetically It is well known that the locus the orthocenter H of of a triangle ABC, when C varies along a parallel g to AB, is a parabola. However, the only proof I am aware of is by coordinatizing the situation.
Does anybody know how to prove this synthetically? It is readily seen that g must be the directrix of the parabola, but how to find the focal point?
 A: Let $M$ be the midpoint of $AB$.  Let $C_0$ be the point on the line $g$ parallel to $AB$ such that $C_0M$ is perpendicular to $AB$.  Let $H_0$ be the orthocenter of $\triangle ABC_0$.  Suppose that $F$ on the segment $H_0M$ such that
$$\overline{H_0F}=\frac{1}{4}\overline{C_0M}$$
and set $F'$ to be the image of $F$ under the reflection about the point $H_0$.  Let $d$ denote the line parallel to $AB$ passing through $F'$.  We claim that $H$ traces the parabola with focal point $F$ and with directrix $d$.

First of all, suppose that $h=\overline{MC_0}$ and $c=\overline{AB}$.  Let $C$ be an arbitrary point on $g$.  Write $x=\overline{CC_0}$ and we attach a sign to $x$ so that $x>0$ if $C$ is closer to $B$ than to $A$, and $x<0$ otherwise.  Let $Z$ be the foot of the altitude from $C$ of $\triangle ABC$.   We note that
$$\overline{AZ}=\frac{c}{2}+x$$
and 
$$\overline{ZB}=\frac{c}{2}-x,$$
where the lengths are signed (positive lengths are oriented the same way as $AB$).  Note that $\triangle AHZ\sim \triangle CBZ$.  That is,
$$\frac{\left|\frac{c}{2}+x\right|}{|\overline{ZH}|}=\frac{|\overline{AZ}|}{|\overline{ZH}|}=\frac{|\overline{CZ}|}{|\overline{ZB}|}=\frac{h}{\left|\frac{c}{2}-x\right|}\,.$$ 
This means
$$\overline{ZH}=\frac{1}{h}\left(\frac{c^2}{4}-x^2\right)=\frac{1}{h}\left(\frac{c^2}{4}-\overline{MZ}^2\right),$$
where positive lengths denote the lengths in the same direction as $MC_0$. Therefore, we have the following parabola equation between $x=\overline{MZ}$ and $y=\overline{ZH}$:
$$\overline{MZ}^2=4\overline{H_0F}\left(\overline{ZH}-\overline{MH_0}\right),$$
as $$\overline{H_0F}=\frac{1}{4}\overline{C_0M}=-\frac{1}{4}\overline{MC_0}=-\frac{h}{4}$$
and
$$\overline{MH_0}=\frac{\overline{AM}\cdot\overline{MB}}{\overline{MC_0}}=\frac{c^2}{4h},$$
since $\triangle AMH_0\sim \triangle C_0MB$.
It can be proven that if $g$ is neither parallel nor perpendicular  to $AB$, the orthocenter $H$ traces a hyperbola.  The two axes of symmetry this hyperbola are parallel to the two angular bisectors of the angles that the line $g$ makes with the line $AB$.

That is, if $g$ is given by the equation $y=mx$ and the line $AB$ is given by the equation $y=-mx$ (here, $|m|\neq 0,1$ so that $g$ is not perpendicular to $AB$), where $A=(a,-ma)$ and $B=(b,-mb)$, then the hyperbola is given by
$$m^2y^2-x^2-(a+b)(1-m^2)(my-x)=ab(1-m^2)(1+m^2).$$
If $g$ intersects the line $AB$ at $A$ or at $B$, say at $A$, then we obtain a degenerate case where $H$ traces a line passing through $B$ perpendicular to $g$.
Note that in the special case where $g$ is perpendicular to the line $AB$ but neither $A$ or $B$ is the intersection of $g$ with the line $AB$, we have another degenerate case where $H$ traces the line $g$ itself.  If $A$ or $B$ is the intersection of $g$ with the line $AB$, then this intersection is the orthocenter of $\triangle ABC$ no matter where $C$ is on $g$.

