<s>[INST] Gaussian curvature of an ellipsoid proportional to fourth power of the distance of the tangent plane from the center? Is it true that the Gaussian curvature of an ellipsoid is proportional to fourth power of the distance of the tangent plane from the center? I can verify that it holds at the places where the major axes intersect the surface. (Mathworld has an equation for the Gaussian curvature, which simplifies at those points.) But verifying that it holds elsewhere seems like it would get ugly.
The motivation for this question is that Lord Kelvin proved that the charge density on a conducting ellipsoid is proportional to the distance of the tangent plane from the center, while McAllister (I W McAllister 1990 J. Phys. D: Appl. Phys. 23 359 doi:10.1088/0022-3727/23/3/016) finds that under certain assumptions, the charge density on a conducting surface is proportional to the fourth root of the absolute value of the Gaussian curvature. However, I think the assumptions of McAllister's result fail for the ellipsoid (actually I only have access to the abstract, so I'm not sure), so it would be nontrivial to learn that this held for the ellipsoid. (The proportionality is definitely not universal. For a pair of conducting spheres that are far apart and connected by a wire, the exponent is not 1/4. For a deep concavity, all of this definitely fails -- you get a a Faraday cage, which excludes the electric field almost completely.)
 [/INST] I did the same thing for the "elliptic hyperboloids," of one sheet or two sheets, see
 http://mathworld.wolfram.com/EllipticHyperboloid.html
where they have some pictures and formulas. There is enough detail, it is not necessary to use the parametrizations. 
I am going to use a fixed number $$ \delta = \pm 1$$ and surface
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \delta = \frac{z^2}{c^2},   $$
or
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} -  \frac{z^2}{c^2} =  - \delta.   $$ 
It is a bit of a mess, but after you put back in all three variables you get
$$  K = \frac{\delta a^6 b^6 c^6}{ \left( b^4 c^4 x^2 + c^4 a^4 y^2 + a^4 b^4 z^2 \right)^2}    $$ 
If we eliminate $z$ as on the website, we get
$$  K = \frac{\delta a^6 b^6 c^6}{ \left(\delta a^4 b^4 c^2 + b^4 c^2 (c^2 + a^2) x^2 + c^2 a^4 (b^2 + c^2) y^2 \right)^2}    $$ 
which uses, and I kept missing factors in this one,
$$ a^4 b^4 z^2 = \delta a^4 b^4 c^2 + a^2 b^4 c^2 x^2 + a^4 b^2 c^2 y^2.   $$
For the distance, fix some point $(x_0, y_0, z_0)$ on the hyperboloid, find the normal vector $ \left(\frac{x_0 }{a^2}, \;  \frac{y_0 }{b^2}, \; -  \frac{z_0 }{c^2} \right)$ using the gradient of the defining equation, write the tangent plane 
$$  \frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} -  \frac{z_0 z}{c^2} =  - \delta.   $$
Find the multiple of the normal vector (leaving the origin) lying in the tangent plane. Oh, the multiplier is
$$  t = \frac{-\delta a^4 b^4 c^4}{ b^4 c^4 x_0^2 + c^4 a^4 y_0^2 + a^4 b^4 z_0^2 }  $$
So the closest point to the origin is that $t$ times the given normal vector, and the distance of that point from the origin is 
$$ d =  \frac{ a^2 b^2 c^2}{ \sqrt{ b^4 c^4 x_0^2 + c^4 a^4 y_0^2 + a^4 b^4 z_0^2} }  $$
where at one point we use $|\delta| = 1.$
Alright, dropping the $0$ subscripts, we get
$$ d^4 =  \frac{ a^8 b^8 c^8}{ \left( b^4 c^4 x^2 + c^4 a^4 y^2 + a^4 b^4 z^2 \right)^2}  $$
and
$$  K = \frac{ \delta \; d^4}{a^2 b^2 c^2}  $$
A few features to notice: the hyperboloid of one sheet has negative Gauss curvature.  The hyperboloid of two sheets has positive Gauss curvature again. The tangent plane, in either case,  never passes through the origin itself.  
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