Points on an ellipse

For a small eccentricity $e$,

\begin{align*} e^2 &= 1-\frac{b^2}{a^2}\\ \theta(t) &= t+\left( \frac{e^2}{8}+\frac{e^4}{16}+\frac{71e^6}{2048} \right) \sin 2t+ \left( \frac{5e^4}{256}+\frac{5e^6}{256} \right) \sin 4t+ \frac{29e^6}{6144} \sin 6t+O(e^{8}) \\ \begin{bmatrix} x \\ y \end{bmatrix} &= \begin{bmatrix} a\cos \theta(t) \\ b\sin \theta(t) \end{bmatrix} \end{align*}

the arclength spacing is approximately equal for uniform spacing of $t\in [0,2\pi]$.

See also my newer post here.

Perhaps the parallelogram method is suited to your purpose? Here it is in action:

(Image by Wikipedia user Cmarm.)

Here is a simple algorithm that distributes points evenly on an ellipse:

1. Sample the ellipse uniformly according to angle using the polar parametrization.

2. Replace each point with the midpoint of its neighbors.
   Project the midpoint ortohogonally onto the ellipse.

3. Repeat step 2 as needed.