"Show that if $P$ is a hyperbolic point, a neighborhood of $P$ in $M \cap T_P M$ is a curve that crosses itself at $P$ and whose tangent directions at $P$ are the asymptotic directions."
This is question 2.2.21c of Shifrin's Differential Geometry (link).
We are being hinted to work in the coordinates $(x, u)$ with $y=ux$.
We first assume $P$ to be the origin, $M$ locally the graph $z=\frac{1}{2}\left(k_1x^2+k_2y^2\right)+\epsilon(x,y)$, where $\lim_{x,y\to0}\frac{\epsilon(x,y)}{x^2+y^2}=0$, $T_P M$ the $xy$-plane ($z=0$), and the $x-$ and $y-$axes are the principal directions at $P$.
When we change coordinates as hinted, we get that the intersection has the equation $\frac{1}{2}x^2(k_1+k_2u^2)+\epsilon(x,ux)=0$, which holds if $x=0$ or if $x\neq0$ and $\frac{1}{2}(k_1+k_2u^2)+\frac{\epsilon(x,ux)}{x^2}=0$.
The condition $x=0$ refers only to the point $(x,y)=(x,ux)=(0,0)$ which is not a curve so we discard it.
We are now assuming $x\neq0$. As we want to study the intersection in a small neighborhood of $P$, we take the limit as $x$ approaches $0$. We get $\frac{1}{2}(k_1+k_2u^2)=0$, i.e. $u = \pm\sqrt{\frac{-k_1}{k_2}}$.
If I understand correctly, we get two values of $u$, each corresponding to the limit of $y/x$ for the points $(x,y)$ of a curve as $x\to0$ (so as the curve gets closer and closer to the origin)? So $u$ would be the slope of the tangent line to the curve at $P$, in the $xy$-plane?
Therfore we get two directions: $v=\left(1,\pm\sqrt{\frac{-k_1}{k_2}}\right)$. We can verify easily that $\mathrm{II}_P(v,v)=0.$
I think it works, but I'm not sure I understand the meaning behind the change of coordinates (I don't know what's blowup) or the limit (we are near $0$, but not at $0$).
Best Answer
Welcome to MSE. Good effort and good job!
The point is that if the curve intersects $x=0$ in the $xu$-plane at the point $(0,u_0)$, then this says that the original curve in the $xy$-plane has slope $u_0$ at the origin. You might be a bit more explicit about why the limit term $\epsilon(x,ux)/x^2$ goes to $0$ as $x\to 0$.
The "blowing-up" to which I referred in the footnote is like working in polar coordinates at the origin. As you know from your calculus experience, polar coordinates $(r,\theta)$ fail to be a coordinate system at the origin. Nevertheless, a $\theta$ value at $r=0$ tells you the angle, hence the slope if you have a curve passing through the origin in the cartesian plane.
You might just do a simple example. Consider the curve $y^2=x^2$ (which we all recognize as a nice X). In the $(x,u)$-coordinates, this becomes $y^2 = (xu)^2 = x^2$, so, discarding $x=0$, we are left with $u^2=1$. This gives the slopes $\pm 1$ for the curve. Of course, this was too obvious. But try the same thing with $y^2=x^2+x^3$, where you don't just get a union of two lines.