= Mean curvature of a catenoid
{title2=$H=0,\quad K=-1/(a^2\cosh^4u)$}
Scale the <catenoid> by its waist <radius> $a>0$ and choose a <unit normal> with outward radial component:
$$
X(u,v)=(a\cosh u\cos v,a\cosh u\sin v,au),\qquad N(u,v)=\frac{(\cos v,\sin v,-\sinh u)}{\cosh u}.
$$
The <hyperbolic cosine> controls the profile $r(z)=a\cosh(z/a)$. Direct <differentiation> yields the <first fundamental form> and <second fundamental form>
$$
I=a^2\cosh^2u\,(du^2+dv^2),\qquad II=a\,du^2-a\,dv^2.
$$
The coordinate <tangent vectors> are orthogonal, so the meridional and circumferential <principal curvatures> are
$$
k_{\mathrm{mer}}=\frac1{a\cosh^2u},\qquad k_{\mathrm{circ}}=-\frac1{a\cosh^2u},\qquad H=0,\qquad K=-\frac1{a^2\cosh^4u}.
$$
The opposite <principal curvatures> cancel at every point, not just at the waist. This is precisely why the <catenoid is a minimal surface>, while its nonzero <second fundamental form> and negative <Gaussian curvature> show that it is not a <plane>.
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