Example 7. The base of a triangle is of fixed length and position, and the height on this side is of fixed length. Find the equation of the locus of its orthocenter.

 untranslated text:
例7. 三角形的某底边长一定, 位置不变, 且这边上的高是定长, 求它的垂心的轨迹方程。 

 translated text:
Example 7. The base of a triangle is of fixed length and position, and the height on this side is of fixed length. Find the equation of the locus of its orthocenter.
Solve: Establish a rectangular coordinate system with the line containing the base $\mathrm{AB}$ of $\triangle \mathrm{ABC}$ as the $x$-axis and the midpoint of $\mathrm{AB}$ as the origin.
As shown in Figure 7, let the orthocenter $\mathrm{H}$ be $(\mathrm{x}, \mathrm{y}), |\mathrm{AB}|=2 \mathrm{a}, |C D|=h$, then $A(-a, 0)$, $B(a, 0), C(t, h)$, where $t$ is a parameter.

Thus, the equation of $\mathrm{CD}$ is $\mathrm{x}=\mathrm{t}$, the slope of $\mathrm{AC}$ is $k_{A C}=\frac{h}{t+a}$, and the slope of $\mathrm{BE}$ is $k_{B E}=-\frac{t+a}{h}$. The equation of line $\mathrm{BE}$ is $y=-\frac{t+a}{h}(x-a)$.
Therefore, the parametric equations of the foot of the perpendicular $\mathrm{H}$ are
$$
\left\{\begin{array}{l}
x=t, \\
y=-\frac{t+a}{h}(x-a) \text {, eliminating } t \text { gives }
\end{array}\right.
$$
$y=-\frac{1}{h}\left(x^{2}-a\right)$ as the required curve equation.