A surface whose all points are hyperbolic

Consider the hyperbolic paraboloid, that is,  the surface $S$ given by the graph of $z=f(x,y)=x^2-y^2$. By the formula of the Gauss curvature for a surface which is a graph, we have $$K=\frac{f_{xx}f_{yy}-f_{xy}^2}{(1+f_x^2+f_y^2)^2}=-\frac{4}{(1+4x^2+4y^2)^2}<0.$$  

Then all points are negative. Consider the point $p=(0,0,0)\in S$, whose tangent plane is the plane $z=0$. Let $c>0$. Then the intersection of $S$ with planes parallel to $T_pS$ is $$S\cap \{z=c\}=\{(x,y,c): x^2-y^2=c\}=\{(x,y,c):\left(\frac{x}{\sqrt{c}}\right)^2-\left(\frac{y}{\sqrt{c}}\right)^2=1\},$$

that is, an ellipse, for each height $c>0$. Similarly, for $c<0$.