Let $X$ be a very general hypersurface of degree $\ge 5$ and $Y$ be an irreducible cubic hypersurface in $\mathbb{P}^3$. It is known that $X \cap H$, where $H$ is a hyperplane, can have at most $3$ nodes. My question is: can $X \cap Y$ have more than $3$ nodes lying on a plane $H$ ?
I guess that is not possible. I have the following argument for that. If possible, let there are at least $4$ nodes of $X \cap Y$ lying on a plane. Choose two of them and consider the line $l$ joining them. Then $l$ is either tangent to $Y$ on lie in $Y$. But a line intersects a cubic curve at at most $3$ points, a contradiction (as $l$ is tangent to $Y$ at two points, $l$ intersects $Y$ at $4$ points counted with multiplicity). Thus $l$ is contained in $Y$. Since $l$ is also tangent $X$ and contained in $H$, $H$ is tangential to $X$ at those points. Choosing another $2$ points and arguing similarly one can see that $H$ is tangent to $X$ at more than $3$ points, a contradiction. Please correct me if I am making any mistake.
