If $r: J \times \mathbb{R} \rightarrow\mathbb{R}^3$, $r(s,t)=c(s)+tv(s)$ is a parametrization of a surface which Gaussian curvature is nonzero, then for each $s_0$ there exist limits of normal vectors $n_+(s_0):=\lim_{t \rightarrow \infty} n(s_0, t)$, $n_-(s_0):=\lim_{t \rightarrow -\infty} n(s_0, t)$ and $n_+(s_0)=-n_-(s_0)$.
I've managed to find a formula for the unit normal vector:
$$n(s,t)=\frac{r_s \times r_t}{|r_s \times r_t|}=\frac{c'(s) \times v(s)+tv'(s) \times v(s)}{\sqrt{\det\mathbb{I}_{r(s,t)}}}$$.
I also know that the gaussian curvature of this surface is negative, because it's not zero and must be nonpositive (i've proved ellier that the gaussian curvature of ruled surfaces is nonpositive). So every point on this surface is hyperbolic, but I'm not sure if it can be useful.
Edit:
Explicity it will be:$$n(s,t)=\frac{c'(s) \times v(s)+tv'(s) \times v(s)}{\sqrt{c'(s) \cdot c'(s)+2tc'(s) \cdot v'(s)+t^2v'(s) \cdot v'(s)- (c'(s) \cdot v(s))^2}}=\frac{\frac{c'(s) \times v(s)}{t}+v'(s) \times v(s)}{\sqrt{\frac{c'(s) \cdot c'(s)}{t^2}+2\frac{c'(s) \cdot v'(s)}{t}+v'(s) \cdot v'(s)- \frac{(c'(s) \cdot v(s))^2}{t^2}}}$$ I've divided the numerator and the denominator through t.