Suppose that every geodesic of a connected surface $M \subset \mathbb{R}^3$ has positive curvature $\kappa > 0$ and is contained in a plane.
a. Prove that every direction $T_p M$ is an eigenvector of for the shape operator $S_p$, for all $p \in M$
b. Prove that $M$ is contained in a sphere.
My starting work is that I know $\alpha $ is geodesic iff $\alpha''(s)$ is colinear to $U(\alpha(s))$ for each $s$ in the domain of $\alpha$ and $U$ is the unit normal to $M$. This then implies that $\alpha''(s) \cdot \alpha'(s) = 0$ and $\alpha''(s) \cdot U'(\alpha'(s)) =0$ as $\alpha'(s), \ U'(\alpha'(s)) \in T_{\alpha(s)} M $. Finally, (I think) $\alpha$ contained in a plane implies that $U(\alpha(s))$ is constant. I would then jump to say that $U'(\alpha'(s)) = 0$. Is this a correct path to take?
For a., I know that I want to show $k_1 = k_2$, where $k_1$ and $k_2$ are the principal directions of $M$ and are the eigenvalues of $S_p$. Is this the
Geodesics have an accelerator (curvature) vector perpendicular to the surface. So the plane must be perpendicular to the surface. Otherwise suppose it is not perpendicular at $p$, the intersection curve has a curvature vector at $p$ that is inside the plane thus not perpendicular to the surface, a contradiction.
Then the Gauss map sends a geodesic to a curve on the unit sphere that is contained in a (parallel) plane. So the tangent direction (of the geodesic) is sent to the same direction. And since this direction is arbitrary, we proved that every direction is an eigenvector of the shape operator.
Since every direction is the eigen direction the eigenvalues (of different directions) must be equal. This shows every point of the surface is umbilical with positive curvature, thus is part of a sphere.