Not all tangents to a plane curve are bitangents I'm struggling on a question from a previous qualifying exam, and I don't see a clean way to do it.
Let $X\subset \mathbb{R}^2$ be a connected, 1-dimensional, real analytic submanifold, not contained in a line. Why does there exist a tangent line to $X$ that not bitangent (tangent to $X$ at more than one point)? 
Out of curiosity, is the statement still true if we drop the condition that $X$ is real analytic (so just smooth)? 
 A: My first instinct was to approach this in terms of analytic curves (say, in the projective plane) and their dual curves. But this is total overkill. It should just be an elementary exercise in the differential geometry of curves, if we make some reasonable assumptions. When $X$ is contained in a line, its curvature $\kappa$ is identically $0$. Real analyticity of $X$ and the assumption that $X$ is not contained in a line will tell you that the zeroes of $\kappa$ must be isolated, and then the proof will work. (I leave it to you to ponder what happens if you just assume $X$ is $C^2$. How horrendous can the $0$-set of the curvature be then?)
It is really a local exercise, proceeding by contradiction. Assume that we arclength parametrize $X$ by $\alpha(s)$ and let $\beta(s)$ be the (locally, a?) "companion point" so that the tangent lines of $\alpha$ and $\beta$ agree. Write $\beta(s) = \alpha(s)+\lambda(s)T(s)$, where $T(s)$ is the unit tangent vector at $\alpha(s)$. Assuming $\lambda$ is differentiable (away from trouble points this should follow from the implicit function theorem), we have
$$\beta'(s) = \alpha'(s) + \lambda'(s)T(s) + \lambda(s)\kappa(s)N(s),$$
where $T(s),N(s)$ give a smoothly-varying orthonormal basis for $\Bbb R^2$. The fact that $\beta'(s)$ is a scalar multiple of $\alpha'(s)=T(s)$ tells us that $\lambda(s)\kappa(s)=0$ for all $s$. So, away from zeroes of $\kappa$ we must have $\lambda = 0$, which means there is no bitangent companion point at all.
EDIT: @JohnHughes's point is well taken. Since this situation cannot occur (bitangencies occur generically only in isolated situations), it is very difficult to have good intuition. Here is a direct approach, again assuming real analyticity to get only isolated zeroes for $\kappa$. Because we want to work with the tangent lines as affine lines, and not as subspaces, it is most natural to work projectively. I can translate the notation I use if necessary.
Given the curve $\alpha\colon I\to\Bbb R^2$ (for an appropriate interval $I$), again assumed to be arclength-parametrized, we consider $f\colon I\to \Bbb R^3$, with $f(s) = (\alpha(s),1)$. To say that $\alpha(t)$ lies on the tangent line at $\alpha(s)$ is to say that $f(s)\wedge f'(s)\wedge f(t) = 0$ (i.e., the three vectors $f(s),f'(s),f(t)$ are linearly dependent). Symmetrically, to say that $\alpha(s)$ lies on the tangent line at $\alpha(t)$ is to say that $f(t)\wedge f'(t)\wedge f(s) = 0$.  So we're interested in pairs $(s,t)$ with $s\ne t$ and $$F(s,t) = \big(f(s)\wedge f'(s)\wedge f(t),f(t)\wedge f'(t)\wedge f(s)\big) = (0,0).$$
(To be a bit more careful, throughout we must choose an identification $\Lambda^3(\Bbb R^3) \cong \Bbb R$. To be concrete, let's set $T(s)\wedge N(s)\wedge (0,0,1)$ equal to $1$.) Now, we just calculate:
$$DF(s,t) = \begin{bmatrix} f(s)\wedge f''(s)\wedge f(t) & f(s)\wedge f'(s)\wedge f'(t) \\ f(t)\wedge f'(t)\wedge f'(s) & f(t)\wedge f''(t)\wedge f(s)\end{bmatrix} = \begin{bmatrix} -\lambda(s,t)\kappa(t) & 0 \\ 0 & \lambda(s,t)\kappa(s)\end{bmatrix},$$
where, at a point $(s,t)$ with $F(s,t)=(0,0)$, we write $\alpha(t) = \alpha(s)+\lambda(s,t)T(s)$. Note that $\alpha(s)\ne\alpha(t)$ tells us that $\lambda(s,t)\ne 0$. So, so long as $\kappa(s)$ and $\kappa(t)$ are both nonzero, $DF(s,t)$ will be nonsingular, which means that — away from zeroes of curvature — $F^{-1}(0,0)$ consists of isolated points, as required. 
