Exercise 5.4.1 in Smith's Invitation to Algebraic Geometry; lines that are tangent to conic are closed subvariety of Gr$(2,3)$ in $\mathbb{P}^2$

Question

I am trying to solve Exercise 5.4.1 in Karen E. Smith's Invatation to Algebraic Geometry:

Fix an irreducible conic $C$ in $\mathbb{P}^2$. Show that the set of lines in $\mathbb{P}^2$ that fail to meet the conic in exactly two distinct points is a closed subvariety of the Grassmannian of all lines in $\mathbb{P}^2$, namely Gr$(2, 3)$.

We work on the field $\mathbb{C}$.

In an earlier course I learned about duality in projective spaces and upon dualization, this statement is trivial. Duals of irreducible conics remain irreducible conics and a line will always meet a conic in two points, counting multiplicity. Failing to meet a conic in two distinct points is hence equivalent to being tangent to it. So, the dual statement sounds:

Show that the set of points in $\mathbb{P}^2$ that are on an irreducible conic is a closed subvariety of the Grassmannian of all points in $\mathbb{P}^2$, i.e., $\mathbb{P}^2$ itself.

Which is obviously true, because a conic is a closed subvariety of $\mathbb{P}^2$.

However in this course, we haven't talked about duality yet. Further, it does not use the Plücker embedding we learned in the section (Section 4 from Chapter 5) that this problem is from. But it does not seem obvious to me how to do this hands on ...

Answer 1

The equation of a tangent line to a smooth curve $V(f)\subset\Bbb P^2$ at a point $[a:b:c]$ is given by $\frac{\partial f}{\partial x}(a,b,c)x+\frac{\partial f}{\partial y}(a,b,c)y+\frac{\partial f}{\partial z}(a,b,c)z=0$. Since each of these partial derivatives are polynomials, and their common vanishing locus has no intersection with $V(f)$ by the smoothness assumption, we get a morphism $V(f)\to (\Bbb P^2)^*$ by sending $[a,b,c]\mapsto [\frac{\partial f}{\partial x}(a,b,c):\frac{\partial f}{\partial y}(a,b,c):\frac{\partial f}{\partial z}(a,b,c)]$ where each point is mapped to its tangent line. Since a smooth curve in $\Bbb P^2$ is irreducible (if it weren't, then it would have a point where its components met, and that would be singular), this means the dual curve of any smooth curve in $\Bbb P^2$ is an irreducible variety - the image of any projectie variety is closed, and the image of any irreducible set is irreducible.

Now recognize that an irreducible conic is smooth (assuming you're working with classical varieties and everything is reduced) and so you can apply the above paragraph.