Please help me to solve the following problem:
There are curves $C$ of degree $3$ and $Q$ of degree $2$ in $\mathbb{C}P^2$. $P_1, \ldots,P_6$ are intersection points of $C$ and $Q$. $T_i$ is a tagent to $C$ at $P_i$, $i \in \{1,\ldots, 6\}$. $Q_i$ is point of intersection of $T_i$ and $C$.
I need to prove that $Q_1, \ldots,Q_6$ are lying on another curve $Q'$ of degree $2$.
My idea:
I think (I am not 100% sure) that I should somehow use: Cayley–Bacharach theorem. Can you please advice how to use it?
Thanks a lot for your help!
This is a straight forward copy of the proof of Thm7.3. Let $D$ be a conic through five of the six points $Q_i$ and $L$ any line through the sixth, not passing though the other points. Let $G$ be the product of the lines $T_i$.
Then by Noether $$Q^2DL=CA+GB$$ now if $r$ and $s$ are the other two points of intersection of $L$ and $C$ then they must be zeros of $GB$ and since they are not zeros of $G$, by choice of $L$ we have that they are zeros of $B$. However $B$ is linear so $B=L$, and we have $$Q^2DL=CA+GL$$ so that gives that $Q^2D$ and $G$ have exactly the same intersection with $C$ and thus $D$ must go through all the points $Q_i$.