Let $C$ be a non-singular curve in the plane, defined over an algebraically closed field $K$ , and given by a polynomial $f(x,y)=0$ of degree $4$ (i.e. $C$ is a quartic). Let $L$ be a (rational) line in the plane $K^2$ . The intersection divisor of $C$ and $L$ is of the form:

$$(L\cdot C)=P_1+P_2+P_3+P_4$$ where $P_i$ , $i=1,2,3,4$ , are points in $C(K)$ . There are five possibilities:

1. The generic position: all the points $P_i$ are distinct.
2. $L$ is tangent to $C$ : there exist indices $1\leq i\neq j\leq 4$ such that $P_i=P_j$ . Without loss of generality we may assume $P_1=P_2$ and $(L\cdot C)=2P_1 + P_3+P_4$ , and $P_3\neq P_4$ .
3. $L$ is bitangent to $C$ when $P_1=P_2$ and $P_3=P_4$ but $P_1\neq P_3$ . It may be shown that if $\operatorname{char}(K)\neq 2$ then $C$ has exactly $28$ bitangent lines.
4. $L$ intersects $C$ at exactly two points, thus $P_1=P_2=P_3\neq P_4$ . The point $P_1$ is called a flex.
5. $L$ intersects $C$ at exactly one point and $P_1=P_2=P_3=P_4$ . This point is called a hyperflex. A quartic $C$ may not have any hyperflex.

Bibliography

1

S. Flon, R. Oyono, C. Ritzenthaler, Fast addition on non-hyperelliptic genus 3 curves, can be found here.