Question

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

\begin{equation} f(p,q)=\widetilde{f}(p,q)=0 \end{equation}

we also have

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

Answer 1

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?

Answer 2

The answer is clearly yes for d=2,3,4 and 6. I am skeptical about larger degrees.

The family of plane curves of degree (at most) d-1 has dimension (d-1)(d+2)/2. Imposing d(d-1)/2 tangency points with the given curve determines a family of curves of degree d-1 tangent to C of dimension at least d-1. For large d, the family of nonreduced solutions will have dimension bigger than d-1, so this gives no information about existence of reduced solutions. However, for d=2,4 and 6 there must be reduced solutions because the nonreduced families of solutions have dimension 0, 2 and 4 respectively.