I'm trying to find the intersection points of two random conics, $C_∞ ,C_0$ by considering the pencil determined by their $4$ intersection points, (which are on general position) $C_t=tC_∞+C_0, t\in ℝ$. (Note: $C_∞$ has been sent to infinity) and finding at least one member which is the union of two lines; the system $C_∞∩C_0$ is equivalent to $C_∞∩C_t$ for all $t$, so we will then have to calculate the intersection of $C_∞$ with this union.
After finding the corresponding matrix of $C_t$, and setting its determinant equal to zero, I've set $ρ$ to be a root of that cubic. Since $C_ρ$ is degenerate, it has at least one singular point and if we set its gradient equal to the zero vector, we have another $3×3$ matrix with $rk<3$, in fact since in my example the rank was $2$, its null-space corresponds to a unique point on the projective plane. (We don't need to prove that this is indeed a point on the conic, it's obviously the case, since the conic is singular.)
Edit: I just now realised that this matrix is twice the matrix of the quadratic form that corresponds to the conic. So, it's always true that the null-space is contained in the conic when it's non-trivial. In fact, the null-space gives us the singular points of the conic.
The problem is, I intuitively knew that I must find the point of intersection of the two lines, but I've forgotten what to do with this information. Maybe I should parametrise the set of all projective lines passing through that point and require that they are tangent to the conic, which ought to give a second degree equation in the parameter, whose solutions are the lines. Or dehomogenise the conic and use the Taylor expansion at the point to find the tangent cone (which equals the conic), then factor the quadratic form in $x,y$. I'm trying to avoid using the dual conic.
I've figured it out, I'm writing it down in case anybody else has this problem. A natural way (up to a choice of an affine patch in $ℙ^2$) to parametrise the set of all lines in the projective plane through a point $[x_0:y_0:z_0]$, which is naturally a projective line $ℙ^1$, is by picking a non-zero co-ordinate, say $z_0$, and considering the corresponding $z=0$ as the line at infinity. Then we have the points $[x:y:0]$ as $[x:y]$ ranges in $ℙ^1$ as the point of intersection of these lines with the line at infinity, so each line is, parametrically, $[s(x_0,y_0,z_0)+t(x,y,0)]$, where $[s:t] \in ℙ^1$, and the entire family is obtained as $[x:y]$ ranges in $ℙ^1$. All we have to do is consider lines that satisfy the equation of the conic, that is, they give zero for all values $s,t$ for some choice of $x,y$ right? But trivially, since $(x_0,y_0,z_0)$ is in the null-space of the matrix of $C$ it is easy to show that it will suffice $v=t(x,y,0)$ satisfying $v^TCv$ for all $t \in ℝ$. But that's the same as finding the points of intersection of the degenerate conic (a union of two lines) with the chosen line at infinity! So if we find the two intersection points with $z=0$ we'd have found two projective points for each line, and it would be a trivial matter to find parametric equations for them, then eventually their corresponding linear forms. So, we set $z=0$ in the equation of $C_ρ$ and we factor the corresponding binary quadratic form in $x,y$, which is as easy as factoring a quadratic polynomial in a single variable. Needless to say, we must exchange $z$ for $x$ or $y$ if that's our non-zero co-ordinate of the intersection point. As a final comment, this is overall an extremely involved process overall, and I should've used Grobner Bases instead. But knowing how to split a degenerate quadratic form into two linear forms has its own value.