I want to show that if a surface is tangent to a plane along a curve , then the points of this curve are either parabolic or planar. (Do Carmo's book problem)
Now what I thought is , since there is a curve tangent to the plane, that means it is in the plane this means that the normal of given curve $n_{\alpha}$ is also in the plane, which tells me it's perpendicular to the normal of the surface $N$ at any point. Then $$k_n = k_{\alpha} \langle n_{\alpha},N\rangle = 0$$
On the other hand I know that $$k_1 \leq k_{n} \leq k_2 $$ That is, any normal curvature will be bounded by the principal curvatures
So now I can discard an elliptic point. But I could still have an hyperbolic point, since any hyperbolic point has negative $k_1$ and positive $k_2$ meaning that it can have a direction with $k_n = 0$. I think I should try to proove 0 is a principal direction, which would solve the issue, but I dont see how.
Also another idea would be to prove that any point cannot have two directions with normal curvature $0$. Since I already proved hyperbolic points have two directions with normal curvature $0$.
So if anyone can give me a tip on how to discard the hyperbolic point I would appreciate it.
By hypothesis, the unit normal field along the curve is constant
$N(t)=(N\circ{\alpha})(t)=N_0$ for all $t$, then $$N'(t)=dN_{\alpha(t)}(\alpha'(t))=0$$
that is to say that $dN_p(v)=0=0v$, therefore $k_1=0$ o $k_2=0$, ie, $K=\det(dN_p)=0$.
These points on the curve are parabolic or flat.