Multiple tangents on a plane curve corresponds to ordinary $r$-fold points on the dual curve Let $X$ be a (smooth, irreducible) curve of degree $d$ in $\mathbb{P}^2_k$ where $k$ is an algebraically closed field of characteristic $0.$ We say that a line of $\mathbb{P}^2_k$ is a multiple tangent of $X$ if it is tangent to $X$ at more than one point. If $L$ is a multiple tangent of $X,$ tangent to $X$ at the points $P_1, \ldots, P_r$ and if none of the $P_i$ is an inflection point, show that the corresponding point of the dual curve $X^*$ is an ordinary $r$-fold point, meaning that it is a point of multiplicity $r$ with distinct tangent directions.  
This is a question from Hartshorne, and while I believe I know how to prove it (with different methods), I don't think is the kind of solution the exercise warrants. Let me briefly outline the methods I know of how to prove this:


*

*Assume that we know that the Gauss map $X \rightarrow X^*$ is birational. Then it follows that $X$ is the normalization of $X^*$ and we can identify the Gauss map with successive blow-ups at singularities of $X^*.$ Then one checks that essentially by definition that the point $L$ is an ordinary $r$-fold point. To know that the Gauss map is birational, one can argue by the biduality theorem for plane curves in characteristic $0.$   

*Use the Lefschetz principle to reduce to when $X$ is a curve over $\mathbb{C}$ and argue using the analytic topology / by geometry.  

*Pass to completions and try to define the Gauss map locally.  


All of these methods seems to me to be somewhat against the spirit of the material that Hartshorne has introduced so far in the book. The first of the above items uses the biduality theorem, which, while not extremely hard to prove, still is quite a lot for this The second is not algebraic at all the last seems somewhat of a stretch.    
I would be very grateful for a purely algebraic solution of this exercise. Let me explain what I mean by a purely algebraic solution. In some sense, it should only use elementary machinery from scheme theory and not go to the analytic category. If completions are neccessary, they can be used but I would prefer something not using that. Further, I would (if possible) want it to be a proof that would hold true in characteristic $p>0$ as well.  
This question has previously been posted to math.stackexchange, but the answers I received was not what I was looking for so I thought I would post it here. I apologize if this is not the right forum for this question, but I am really curious.
 A: You should really try to do this exercise yourself.  Here is a hint.  Because the characteristic is $0$, by generic smoothness the map from the curve $X$ to the dual projective space has nonzero derivative except at finitely many points.  Thus, for a general point $p$ of $X$, the tangent line $L$ to $X$ at $p$ is not an inflection line, and the same holds for any other point $q$ of $X$ such that $L$ is the tangent line to $X$ at $q$.  
Now choose homogeneous coordinates $[s,t,u]$ on $\mathbb{P}^2$ such that $L$ equals $\text{Zero}(u)$, such that $p$ equals $[1,0,0]$ and such that $q$ equals $[0,1,0]$.  Then $\tau=t/s$ is a local coordinate on $X$ near $p$, and $\sigma=s/t$ is a local coordinate on $X$ near $q$.  Since $L$ is not an inflection line near $p$ nor $q$, locally near $p$, $u/s = \tau^2 a$ as regular sections of $\mathcal{O}_X$, where $a\in \mathcal{O}_{X,p}$ is invertible.  Similarly, $u/t=\sigma^2 b$, where $b\in \mathcal{O}_{X,q}$ is invertible.  Of course the completion $\widehat{\mathcal{O}}_{X,p}$ is isomorphic to $k[[\tau]]$, and similarly $\widehat{\mathcal{O}}_{X,q}$ is isomorphic to $k[[\sigma]]$.  Thus, it makes sense to consider $a$ as a power series $a(\tau) = a_0 + a_1\tau + a_2\tau^2 +\dots$, and similarly for $b$.  
Near $p$, on an open neighborhood where $\tau$ is a coordinate and where $a$ is invertible, the equation of the tangent line at $\tau_0$ is $$(u-\tau_0^2 a(\tau_0)s) - (2\tau_0 a(\tau_0) + \tau_0^2 a'(\tau_0))(t-\tau_0 s) =
$$
$$
\tau_0^2( a(\tau_0) - \tau_0 a'(\tau_0))s + \tau_0(-2a(\tau_0) - \tau_0a'(\tau_0))t + u.$$  In particular, if we represent the universal line in $\mathbb{P}^2$ as $$As + Bt + Cu = 0,$$ so that the homogeneous coordinates in the dual projective space are $[A,B,C]$, then the tangent line to the dual curve of $X$ near $p$ is given by $\text{Zero}(A)$.  Now, by symmetry, do the same computation for $q$.  You should have that the tangent line near $q$ is given by $\text{Zero}(B)$.  Thus, if $L$ is not an inflection line to $X$, then even if it happens that two points $p$ and $q$ both map to $[L]$ under the dual curve morphism, nonetheless, the corresponding branches of the dual curve at $[L]$ have distinct tangent lines.
Notice, this goes wrong in characteristic $2$, because then $-2a(\tau_0)$ is zero.  Also, in every positive characteristic $p$, there does exist a "funny curve" such that every tangent line to $X$ is an inflection line.  However, in characteristics $p\neq 2$, so long as $X$ is not a "funny curve", then the analysis above is correct.
