Elliptic and parabolic points.

Two days ago, we have shown a surface where all its points are hyperbolic. Now, we consider the elliptic paraboloid $z=x^2+y^2$. Now $$K=\frac{4}{(1+4x^2+4y^2)^2},$$proving that all its points are elliptic. 

On the other hand, the parabolic paraboloid $z=x^2$ has as Gauss curvature $K=0$ so all its points are parabolic or flat. In order to distinguish, we have to compute the principal curvatures. Since $K=0$, the principal curvatures are $\kappa_1=H$ and $\kappa_2=0$ by the relation $$\kappa_i=H\pm\sqrt{H^2-K}.$$ The mean curvature is given by $$H=\frac12\frac{(1+f_y^2)f_{xx}-2f_xf_yf_{xy}+(1+f_x^2)f_{yy}}{(1+f_x^2+f_y^2)^{3/2}}=\frac{1}{(1+4x^2)^{3/2}}\not=0$$ for any $x$. This proves that all its points are parabolic.

In the next pictures we have a elliptic paraboloid (left) and a parabolic paraboloid (right).