Let $\gamma\colon I\to S$, where $I$ is an interval and $S$ is a surface then the curvature restricted to $\gamma$ is $K(\gamma(t)))=0$ in the parabolic set $\gamma$, if we derive the \begin{equation*} dK_{(\gamma(t)}\cdot(\gamma'(t))=0 \end{equation*} If $X\colon U\to S$ then $\gamma(t)=X(u(t),v(t))$, this implies that $\gamma'(t)=X_{u}u'+X_{v}v'$, characterizing the tangent vector $\gamma'(t)$ in terms of $dK_{(\gamma'(t))}$ we have: \begin{align*} \nabla K_{\gamma(t)}\cdot\gamma'(t)&=0\\ (k_u,k_v)\cdot\gamma'(t)&=0\\ (k_u,k_v)\cdot(-k_v,k_u)&=0 \end{align*} Assuming the parametrization is orthogonal. To verify that $\gamma'(t)$ is an asymptotic line when the normal curvature \begin{align*} K_{n}(\gamma'(t))&=0\\ \end{align*} that is, it is an asymptotic direction when the second fundamental form is zero.
I need help or suggestion on the following: I should show that in ordinary parabolic points $k_{n}(\gamma'(t))\neq 0$, that is, it is not an asymptotic direction.
In an ordinary parabolic point the direction is transversal to the set, which corresponds to a singularity of type fold.
If we want the point to be more generated, show that the direction is tangential to the set, which corresponds to a singularity of type cusp.
I have tried but could not solve it, it would be helpful to know what conditions or how to show this.