intersection divisor for a quartic

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 $1\le i\ne j\le 4$ 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 $P_3\ne P_4$.

3. $L$ is bitangent to $C$ when $P_1=P_2$ and $P_3=P_4$ but $P_1\ne P_3$. It may be shown that if $\mathrm{char}(K)\ne 2$ then $C$ has exactly $28$ bitangent lines.

4. $L$ intersects $C$ at exactly two points, thus $P_1=P_2=P_3\ne P_4$. 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.