Parabola or Catenary in this case? Exhibit A: the flexible film sinks into the box due to lower internal pressure inside the box. 
question is, does the film form a paraboloid or a 3D catenary or neither? this is the usual method used to make low cost reflective parabolic surfaces (mylar film). problem is that parabolic surfaces made this way will probably work satisfactorily even if they weren't technically parabolic. I want to know if this method of using air pressure can truly, in principle,create a parabolic surface, or does it just create an approximation, and if the latter, what would be the mathematical model?
the hanging chain affixed at two ends example is usually touted for catenaries. the main gist I get from this is that all force vectors experienced by the arch are tangent to the arch, everywhere on the arch.. 
Meanwhile the arched cable in suspension bridges is the example of choice for parabolas. in this case the force vector at the vertex is perpendicular to the arch and all external forces acting on the arch are parallel to this.
In exhibit A, due to the nature of air pressure, the external force vectors are perpendicular to the surface everywhere on the surface. I suspect that the curve model is spherical instead of parabolic.
any input is valuable!

 A: This is a well studied problem. The classical solution is due to Henckey:
 "On the stress state in circular plates with vanishing bending stiffness",
 Zeitschrift für Mathematik und Physik, vol 63, 1915, pp 311-317; there was a follow-up publication by NASA that shows the original derivation breaks down at larger deflections, when the radial component of stress is no longer negligible. They also correct an algebraic error in one of the terms (they didn't have Mathematica in 1915, and some of these expressions get pretty hairy...)
It turns out that the shape is "approximately parabolic", but if you read the above paper you will see that the actual shape is a function of the Poisson ratio of the material. In particular, the general solution is of the form
$$W(\rho)=q^{1/3}\sum_0^{\infty} a_{2n}\left(1-\rho^{2n+1}\right)$$
Where $\rho = r/a$ is the dimensionless distance from the center of the circle, $q$ is a dimensionless loading parameter $q=\frac{pa}{Eh}$, $a$ is the radius of the membrane, $h$ the thickness, $E$ the Young's modulus and $p$ the applied pressure.
The values of $a_2n$ depend on $\mu$, the Poisson ratio. They are expressed in terms of $b_{2n}$, another set of coefficients used to compute the stress. The first few terms are
$$\begin{align}a_0 &= \frac{1}{b_0}\\
a_2 &= \frac{1}{2 b_0^4}\\
a_4 &= \frac{5}{9 b_0^7}
\end{align}$$
More terms can be found in the NASA paper. The values of $b_0$ are tabulated as a function of $\mu$ in the same paper:
 mu     b_0
0.2    1.6827
0.3    1.7244
0.4    1.7769

As you can see, the value doesn't change "very much". If we use the values for $\mu=0.4$, then the form of the membrane becomes
$$W(\rho) = q^{1/3} \left(0.5628(1-\rho^2) + 0.0502(1-\rho^4) + 0.0099(1-\rho^6)\right)$$
I plotted this expression (normalized to a deflection of w=1 at $\rho=0$), as well as a pure quadratic. You can clearly see that there are deviations from the quadratic shape:

Incidentally, you may find that "Modeling large deflection of circular membranes for applications in active optical elements", Novàk et al, 2014, doi:10.1117/12.2061039 (behind a paywall) gives you good additional information - it is specifically talking about using such membranes as optical elements, and studies in particular the deviations from ideal behavior for larger deflections. I was not able to read beyond the abstract though, so I can't be sure...
