TITLE: Can the two branches of a hyperbola have more than one common normal?
QUESTION [5 upvotes]: Today I learned that the two branches of the standard hyperbola $\displaystyle\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ have no common tangents, but have only one common normal ($y=0$). So I wondered if if it has more than one common  normal as well.
I began by taking two parametric points $P(\theta_1)$ and $Q(\theta_2)$. We know that a general slope of normal is given by: $-\frac ab\sin\theta$. Setting slope on both points equal we get: $\sin\theta_1=\sin\theta_2$. This implies either $\theta_1=\theta_2=0$ (a case which we already covered) or $\theta_1+\theta_2=\pi$.
However, a quick desmos graph shows that the normals I've calculated are instead parallel and non-intersecting: (for $\theta_1=30^\circ$ and $\theta_2=150^\circ$)

So, is it true that the two branches of a hyperbola have not more than one common normal? Or did I miss some calculation in my step?

REPLY [3 votes]: Intuitive: take a point $P$ on the left piece, below the horizontal. Take the normal to that branch at that point. It will point downwards, so if it intersects the right piece, it will be a point lower $P'$. Now  take the normal to the right piece at that point $P'$. It will go lower from right to left. So if it intersects the left piece, it will be at a point $P''$ lower than $P'$, and so lower than $P$, so cannot get back to $P$.
In general, if we have two convex closed regions with  smooth boundary that are disjoint then they have have a common normal $AB$. Indeed, we can consider points $A$, $B$ in each of the regions that are closest.  Assume moreover that the line $AB$ intersects the each region along a half line and at least one of the region is strictly convex. Then $AB$ is the only common normal. The proof is the same.