I was trying to solve the following problem:
Show that if a surface is tangent to a plane along a curve, then the points of this curve are either parabolic or planar.
I found the following solution:
if $N$ is the Gauss map, and $\alpha(t)$ is the curve, then $N(\alpha(t))$ is normal to the plane and hence constant. Thus $0 \equiv dN/dt = dN(\alpha'(t))$ at all points of the curve. Assuming $\alpha$ is a regular parametrization, $\alpha'(t) \neq 0$, and so $\ker dN \neq 0$, i.e $dN$ is not injective and $det(dN) \equiv 0$ on $\alpha$. The result follows by definition.
My question is: why do we need to prove that $dN$ is not injective? Once we prove that $0 \equiv dN/dt$, we are basically done since this directly implies that $\det(dN) \equiv 0$ on $\alpha$.
You do not need to analyze injectivity. You just need to find a vector $v$ such that $dN_p(v) = 0$ because if this is the case, then $0$ is an eigenvalue of $dN_p$ which implies that $\det dN_p = 0$. Then all the points on the curve are either parabolic or planar by definition.