If all points of a coordinate nbhood are parabolic, then the unique asymptotic curve is a straight line 
Do Carmo's exercise: Let $p$ be a point of an oriented surface $S$ and assume that there is a neighborhood of $p$ in $S$ all points of which are parabolic. Prove that the (unique) asymptotic curve through $p$ is an open segment of a straight line.

At the end of the book the following hint is given:

Parametrize a neighborhood of $p\in S$ in such a way that the lines of curvature are the coordinate curves and that $v= const. $ are the asymptotic curves. It follows $e_v=0$, and from the Mainard-Codazzi equations, we conclude that $E_v= 0$. This implies that the geodesic curvature of $v=const.$ is zero.

Since all the points are not umbilical, I know that we may parametrize a neighborhood of $p$ in such a way that the coordinate curves are lines of curvature. I can't see though why we can make $v=const.$ be also an asymptotic curve. Assuming that we may do this, I also can't see why it implies $e_v=0$. The remaining affirmations I agree, after all the coordinate curves being lines of curvature implies $f=F=0$ and hence from the Mainard equations we have $0 = e_v=E_v/2(e/E+g/G)$ which implies $E_v=0.$ Since the geodesic curvature of the $u$-coordinate curve is $-E_v/(2E\sqrt G),$ it follows that the coordinate curve is geodesic and asymptotic simultaneously. Therefore, by another exercise in this same section, we conclude that it is a segment of a line.
Any help in clarifying these claims is much appreciated
 A: Let $U$ be the neighborhood of $p$ which exists a parametrization where the coordinate curves are lines of curvature and all points are parabolic.
Then, $k_1 k_2 = \det (dN_q) = 0 \; (\forall q \in U)$, where $k_1, k_2$ are principal curvatures.
$\forall q \in U$, since $dN_q \neq 0$, one of $k_1, k_2$ is zero and the other is not.
Since the coordinate curves are lines of curvature in $U$, w.l.o.g., assume $x_u, x_v$ are principal directions corresponding to $k_1, k_2$, respectively. Note that principal directions are orthogonal.
W.l.o.g., assume $k_1 = 0$ at $p$, therefore $x_u$ is the only asymptotic direction of $p$ (since $k_2 \neq 0$).
$$N_u = dN_p(x_u) = k_1 x_u = 0 $$
$$N_v = dN_p(x_v) = k_2 x_v \neq 0 $$
Since $S$ is $C^\infty$, $N_v$ is continous on $U$, therefore there exists a neighborhood $V \subset U$ such that $N_v(q) \neq 0 \; (\forall q \in V)$.
Therefore $N_u = 0 \; (\forall q \in V)$.
Thus, $v=const.$ are the asymptotic curves in such neighborhood $V$ (In other words, all curves $C \subset V$ of the form $v=const.$ are asymptotic).
Moreover, $e = - \langle N_u, x_u \rangle = 0 \; (\forall q \in V)$. Hence $e_v = 0$.
A: Correct me if I am wrong, the following notations agree with those in Do Carmo's textbook.
Since $u=const.$ is line of curvature and $v=const.$ is asymptotic curve, it follows $X_u$ is asymptotic direction, and $X_v$ is principle direction, therefore $N_v=\lambda X_v$ for some real valued function $\lambda$, also $II(X_u)=\langle -dG(X_u),X_u\rangle=\langle -N_u,X_u\rangle$=0. By definition, $e=\langle N,X_{uu}\rangle$, thus
\begin{align*}
e_v&=\langle N_v,X_{uu}\rangle+\langle N,X_{uuv}\rangle=\lambda\langle X_v,X_{uu}\rangle-\langle N_u,X_{uv}\rangle\\
&=\lambda\langle X_v,X_{uu}\rangle-\langle N_u,X_u\rangle_v+\langle N_{uv},X_u\rangle\\
&=\lambda\langle X_v,X_{uu}\rangle-(0)_v+\langle (\lambda X_v)_u,X_u\rangle\\
&=\lambda\langle X_v,X_{uu}\rangle+\lambda\langle X_{uv},X_u\rangle+\lambda_u\langle X_v,X_u\rangle\\
&=\lambda(\langle X_v,X_{uu}\rangle+\langle X_{uv},X_u\rangle)+\lambda_uF=\lambda F_u+\lambda_u F=0
\end{align*}
Then the result follows by the op's argument.
