Cubic hypersurface of singular conics Conics in $\mathbb{P}^2$ are in one to one correspondence with points in $\mathbb{P}^5$, simple enough. Conics of rank one i.e. double lines are in a one to one correspondence with points on the Veronese surface in $\mathbb{P}^5$, this is also clear to me. Now I read that all singular conics in $\mathbb{P}^2$ (i.e. rank 1 or 2) somehow correspond to a cubic hypersurface in $\mathbb{P}^5$. I also read that this cubic hypersuface has something to do with the secants lines of the Veronese surface, which I also don't follow. Could you clarify this last correspondance for me?
 A: The space of conics in $\mathbf P^2$ can be thought of as the space of symmetric $3 \times 3$ matrices $M$, modulo scalars, and the rank of a conic $C$ is exactly the rank of a corresponding matrix $M$. (There is a 1-parameter family of such $M$ for a given $C$, but they are all multiples of each other, so the rank is well-defined.) 
So the hypersurface of singular conics is exactly 
$$ \{ \operatorname{det} M =0 \} \subset \mathbf P^5.$$
The determinant of a $3 \times 3$ matrix is a homogeneous cubic polynomial in the entries, so this is a cubic hypersurface.
Now for the second part: a point on the secant variety of the Veronese surface can be thought of as a linear combination of two matrices of rank 1 (again, modulo scalars). It's not hard to see that such a matrix has rank at most 2. So the secant variety of the Veronese surface is contained in the determinant hypersurface. On the other hand, one can show that any  matrix of rank 2 is a linear combination of two rank-1 matrices (say by using projective transformations to diagonalise the rank-2 matrix). So actually the secant variety is equal to the determinant hypersurface. 
