The vertex C of the triangle ABC is allowed to vary along a line parallel to AB. Find the locus of the orthocenter. 
1. **Set up the coordinates:**
   We place the points \( A \) and \( B \) on the x-axis for simplicity. Let \( A = (-1, 0) \) and \( B = (1, 0) \). The point \( C \) is allowed to vary along a line parallel to \( AB \), so we can write \( C = (m, k) \) where \( k \) is a fixed real number.

2. **Determine the x-coordinate of the orthocenter:**
   Since \( C \) varies along a line parallel to \( AB \), the x-coordinate of the orthocenter will be the same as the x-coordinate of \( C \). Therefore, the x-coordinate of the orthocenter is \( m \).

3. **Find the equation of the line \( BC \):**
   The slope of \( BC \) is given by:
   \[
   \text{slope of } BC = \frac{k - 0}{m - 1} = \frac{k}{m - 1}
   \]
   Thus, the equation of the line \( BC \) is:
   \[
   y = \frac{k}{m - 1}(x - 1)
   \]

4. **Find the equation of the altitude from \( A \) to \( BC \):**
   The slope of the altitude from \( A \) to \( BC \) is the negative reciprocal of the slope of \( BC \):
   \[
   \text{slope of altitude} = -\frac{m - 1}{k}
   \]
   The equation of the altitude from \( A \) is:
   \[
   y = -\frac{m - 1}{k}(x + 1)
   \]

5. **Find the y-coordinate of the orthocenter:**
   The orthocenter is the intersection of the altitudes. We need to find the y-coordinate of the orthocenter. The altitude from \( B \) to \( AC \) has a similar form, but we can use the fact that the orthocenter lies on the altitude from \( A \):
   \[
   y = -\frac{m - 1}{k}(m + 1)
   \]
   Simplifying this, we get:
   \[
   y = -\frac{m^2 - 1}{k}
   \]
   Since \( m \) is the x-coordinate of the orthocenter, we replace \( m \) with \( x \):
   \[
   y = -\frac{x^2 - 1}{k}
   \]

6. **Determine the locus of the orthocenter:**
   The locus of the orthocenter is given by the equation:
   \[
   y = -\frac{x^2 - 1}{k}
   \]

The final answer is \(\boxed{y = -\frac{x^2 - 1}{k}}\)