You link has broken in the last six years, but something like that expression now appears in Annex B (page 47) of https://www.ordnancesurvey.co.uk/docs/support/guide-coordinate-systems-great-britain.pdf which has $\nu =\frac{ a} {\sqrt{1 - e^2 \sin^2(\varphi) }}$
If the world was spherical with radius $r$ then a point with latitude $\varphi$ and longitude $\lambda$ would have $(X,Y,Z)$ co-ordinates
- $X=r \cos(\varphi) \cos(\lambda)$
- $Y=r \cos(\varphi) \sin(\lambda)$
- $Z=r \sin(\varphi)$
but with an ellipsoid, the coordinates of a point on the ellipsoid (ignoring height above the ellipsoid) would be
- $X=\nu \cos(\varphi) \cos(\lambda)$
- $Y=\nu \cos(\varphi) \sin(\lambda)$
- $Z=(1-e^2)\nu \sin(\varphi)$
with $\nu$ defined as above. Note that $\nu$ varies with latitude $\varphi$
So $\nu$ is close to the idea of a radius: because of the $(1-e^2)$ term, it is not quite the distance from the centre of the ellipsoid to a point of latitude $\varphi$, but it is not far away
Personally I would just call it nu
or perhaps something like ellipsoid_nu