Starting with an ellipse, defined with the positions of the foci, and with the major axis length (the semi-major axis is half of the length of the string for the pin and string method), I am trying to create a map between an x-value (which I would like to be normalised at the end) and a point on the ellipse.
For example, if I specify a value of $\frac{1}{2}$, I would like for that to give me the cartesian coordinate of the location on the ellipse that is $\frac{1}{2}$ of the way across the ellipse.
To be more clear, in an ellipse where the foci are at $(\pm 1,0)$, and the semi-major axis is $\sqrt{2}$, (Which makes the semi-minor axis 1), a value of zero should return $(\sqrt{2},0)$, $\frac{1}{2}$ should give $(0,-1)$, etc.
I am looking for a general formula that will provide a continuous solution for the problem (of course, you need only worry about either the bottom or the top branch of the ellipse, whichever you prefer - I will use the bottom (or negative-y)).
You can write any ellipse (centered in $0$, with foci either both on the $x$ or the $y$-axis, otherwise translate/rotate) in the form
$$\left\{(x,y)\in \mathbb{R}^2: \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1\right\}$$ where $a$ and $b$ are the semi-axes. But then
$$y = \pm b\sqrt{1-\left(\frac{x}{a}\right)^2}$$
which is enough for your example. If you want arbitrary foci then you'll need to add some affine transformation $(x,y) \mapsto R(x,y) - c$ where $c$ is the center of the ellipse and $R$ is a rotation by minus the angle between the coordinate axes and the axes of the ellipse. This will map the your ellipse on the one above. Then you end up with
$$\left(\frac{R(x,y)-c}{a}\right)^2 + \left(\frac{R(x,y)-c}{b}\right)^2 = 1$$
which is still a polynomial of degree 2, so you can solve for y. (However I won't write down the whole calculation, which looks rather boring and ugly.)