# Confusion on the relationship between asymptotic directions and $dN_p$ for a point on a surface

I am using Do Carmo's Differential Geometry of Curves and Surfaces textbook and I am a little confused on section 3.2. An asymptotic direction of S at p is defined to be a direction for which the normal curvature is zero. Exercise 1 of 3-2 implies that hyperbolic points have asymptotic directions. Exercise 2 asks to show that if a surface is tangent to a plane along a curve, then the points of this curve are either parabolic or planar, and the solution makes use of the fact that $$dN_p(v) = 0$$ implies $$det[dN_p]=0$$. (Thus, point is parabolic or planar)

The part where I am confused about is I thought that the normal curvature equalling zero along some direction meant that the normal vector would stay constant along that direction, and thus $$dN_p(v)$$ would be 0. But, since hyperbolic points have points where the normal curvature is 0, this would invalidate the solution for exercise 2. So, is my understanding of normal curvature and differential of the Gauss map flawed? On a surface $$S$$ at a point $$p$$ with a direction $$v$$, how does the normal curvature at $$p$$ along $$v$$ relate to $$dN_p(v)$$ if at all?

Solution for exercise 2 found here: http://mathhelpforum.com/differential-geometry/57838-parabolic-planar-points.html

Sorry for being quite wordy, as this is my first question asked here and I wasn't quite sure how to phrase it.

Welcome to MSE. Normal curvature in the direction of a unit vector $$v$$ is $$\pm \langle dN_p(v),v\rangle$$. So normal curvature $$0$$ merely signifies that $$dN_p(v)$$ is orthogonal to $$v$$, not that it is $$0$$ itself. A good example is to consider a helicoid along one of the rulings: The tangent plane twists as you move along the ruling. (You might enjoy reading my differential geometry text as an additional resource.)