# Algebraic plane curves and their tangent curves

Given an algebraic plane curve how can one construct a corresponding curve whose degree is one less and such that the points of intersection are also points of tangency?

Specifically, if a plane curve $F$ is given by the polynomial equation $f(x,y) \equiv \sum_{j=0}^{n}\sum_{i=0}^j{a_{i,j}x^{j-i}y^i}=0$, where $x,y \in \mathbb{C}$, how can one construct a corresponding plane curve $\widetilde{F}$, given by $\widetilde{f}(x,y) \equiv \sum_{j=0}^{n-1}\sum_{i=0}^j{\widetilde{a}_{i,j}x^{j-i}y^i}=0$ such that for a pair $p,q$ satisfying

$$f(p,q)=\widetilde{f}(p,q)=0$$

we also have

$$\dfrac{\partial_xf}{\partial_x \widetilde{f}} \bigg|_{x=p,y=q}=\dfrac{\partial_yf}{\partial_y \widetilde{f}} \bigg|_{x=p,y=q}$$?

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I probably do not understand what you mean: but why doesn't $\tilde F(x,y)=1$ work? –  Mariano Suárez-Alvarez May 6 '11 at 21:26
Because that polynomial has no roots, i.e. $\widetilde{f}(x,y) \equiv 1 \neq 0$ and does not constitute an algebraic plane curve –  hailekofi May 6 '11 at 21:39

If $n$ is odd, you can take an arbitrary $g$ of degree $(n-1)/2$ and $\tilde f = g^2$. The solution is far from unique and there will be others. For $n=2$, $\tilde f$ is the tangent line at some point of $F$. For $n=3$, if you have a smooth cubic with inflexion $O$, taken as origin of the group law, choose a point of order $2, P_0$. For arbitrary points $P,Q$ on the curve, let $R= -(P+Q)+P_0$. Then $2(P+Q+R)=0$ so there is a conic through $P,Q,R$ tangent to $F$ at these points. The case $P_0=O$ reduces to my first construction, but as you see there are three other families. Do you have any reason to believe there will be a nice formula?
You don't need n to be odd for the nonreduced trick: let g have degree one and take $\tilde f=g^{d-1}$. I guess the interesting question is whether a nonreduced all-tangent curve exists. A quick parameter count makes me skeptical... –  quim May 7 '11 at 8:17