Intersection of a smooth cubic surface with a plane We know that the intersection of a smooth cubic surface in $\mathbb{P}^3$ with a plane is a cubic, a line+a conic or three disjoint lines. Can this intersection consist of a conic and its tangent line?
 A: I guess so. Try the cubic surface whose equation is $$x(xy-z^2)-y^3+w^3=0.$$
If my computations are correct, it is smooth and the intersection with any of the three planes $y- \eta w=0$ (where $\eta$ is a third root of unity) is isomorphic to  $x(xy-z^2)=0$, which is the union of a smooth conic and one of its tangent lines.
A: In fact, the result is more general. Let $S$ be any smooth cubic in $\mathbb{P}^3$ and let $L\subset S$ be any line (there are $27$ on it). Then, there exist exactly two conics $C_1,C_2$ in $S$ which are tangent to $L$ and such that the union of $L$ with $C_i$ is the intersection of $S$ with a plane. 
To show this, let $\pi\colon S\to \mathbb{P}^1$ be the map given by the projection from the line (for example if $L$ has equation $x=y=0$, take $\pi\colon (w:x:y:z)\mapsto (x:y)$). One can see that the map $\pi$ is a morphism (for $x=y=0$, the equation of $S$ is $xL=yM$ where $L,M$ are two polynomials of degree $2$ and you send $(w:x:y:z)$ on $(M:L)$).
The fibres of $\pi$ correspond to the conics of $S$ such that the union with $L$ is the intersection of $S$ with a plane. In general, such a conic intersects $L$ into $2$ points, so $\pi$ induces a $2:1$ map from $L$ to $\mathbb{P}^1$. By Hurwitz-formula, the map has exactly two ramification points, which are the two conics meeting $L$ in only one point, i.e. with a tangence.
PS: Cutting $S$ with a plane, it is also possible to get a singular irreducible cubic, which can be cuspidal or nodal. Or three lines. A similar argument shows that you can choose any $L$ and find five pairs of $2$ lines associated to it.
PS2: It was not a stupid post. Even if the answer is easy, it gives nice questions of elementary geometry.
