Problem: For a point $P = (a,a^2)$ in the coordinate plane,
let $\ell(P)$ denote the line passing through $P$ with slope $2a$.
Consider the set of triangles with vertices of the form
$P_1 = (a_1, a_1^2)$, $P_2 = (a_2, a_2^2)$, $P_3 = (a_3, a_3^2)$,
such that the intersection of the lines $\ell(P_1)$, $\ell(P_2)$, $\ell(P_3)$
form an equilateral triangle $\Delta$.
Find the locus of the center of $\Delta$ as $P_1P_2P_3$ ranges over all such triangles.

Solution: The answer is the line $y = -1/4$.
I did not find this problem inspiring,
so I will not write out most of the boring calculations
since most solutions are just going to be
``use Cartesian coordinates and grind all the way through''.

The ``nice'' form of the main claim is as follows
(which is certainly overkill for the present task,
but is too good to resist including):
\begin{claim*}
  [Naoki Sato]
  In general, the orthocenter of $\Delta$ lies on
  the directrix $y = -1/4$ of the parabola
  (even if the triangle $\Delta$ is not equilateral).
\end{claim*}
\begin{proof}
  By writing out the equation $y = 2a_i x - a_i^2$ for $\ell(P_i)$,
  we find the vertices of the triangle are located at
  \[
    \left( \frac{a_1+a_2}{2}, a_1a_2 \right); \quad
    \left( \frac{a_2+a_3}{2}, a_2a_3 \right); \quad
    \left( \frac{a_3+a_1}{2}, a_3a_1 \right).
  \]
  The coordinates of the orthocenter can be checked explicitly to be
  \[ H = \left( \frac{a_1 + a_2 + a_3 + 4 a_1 a_2 a_3}{2},
      - \frac 14 \right). \]
  An advanced synthetic proof of this fact is given at
  \url{https://aops.com/community/p2255814}.
\end{proof}

This claim already shows that every point lies on $y = -1/4$.
We now turn to showing that, even when restricted to equilateral triangles,
we can achieve every point on $y = -1/4$.
In what follows $a = a_1$, $b = a_2$, $c = a_3$ for legibility.

\begin{claim*}
  Lines $\ell(a)$, $\ell(b)$, $\ell(c)$
  form an equilateral triangle if and only if
  \begin{align*}
    a+b+c &= -12abc \\
    ab+bc+ca &= -\frac34.
  \end{align*}
  Moreover, the $x$-coordinate of the equilateral triangle is $\frac13(a+b+c)$.
\end{claim*}
\begin{proof}
  The triangle is equilateral if and only if the centroid and orthocenter coincide, i.e.
  \[ \left( \frac{a+b+c}{3}, \frac{ab+bc+ca}{3} \right) = G = H =
    \left( \frac{a+b+c+4abc}{2}, -\frac14 \right). \]
  Setting the $x$ and $y$ coordinates equal,
  we derive the claimed equations.
\end{proof}

Let $\lambda$ be any real number.
We are tasked to show that
\[ P(X) = X^3 - 3\lambda \cdot X^2 - \frac34 X + \frac{\lambda}{4} \]
has three real roots (with multiplicity);
then taking those roots as $(a,b,c)$
yields a valid equilateral-triangle triple
whose $x$-coordinate is exactly $\lambda$, be the previous claim.

To prove that, pick the values
\begin{align*}
  P(-\sqrt{3}/2) &= -2\lambda \\
  P(0) &= \tfrac14 \lambda \\
  P(\sqrt{3}/2) &= -2\lambda.
\end{align*}
The intermediate value theorem
(at least for $\lambda \neq 0$) implies that $P$
should have at least two real roots now,
and since $P$ has degree $3$, it has all real roots.
That's all.