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I am currently in a problem in the context of my PHD where I need to find the intersection points of two conics $C_A$ and $C_B$.

I do not have direct access to those two conics themself, but rather a handful of 6 conics $C_1$, $C_2$, $C_3$, $C_4$, $C_5$ and $C_6$ of the intersection's pencil, so that

\begin{matrix} C_1=a_A C_B - a_B C_A=& &+ b_1 y^2 &+ c_1 xy &+ d_1 x &+ e_1 y &+ f_1 \\ C_2=b_A C_B - b_B C_A=& a_2 x^2 & &+ c_2 xy &+ d_2 x &+ e_2 y &+ f_2 \\ C_3=c_A C_B - c_B C_A=& a_3 x^2 &+ b_3 y^2 & &+ d_3 x &+ e_3 y &+ f_3 \\ C_4=d_A C_B - d_B C_A=& a_4 x^2 &+ b_4 y^2 &+ c_4 xy & &+ e_4 y &+ f_4 \\ C_5=e_A C_B - e_B C_A=& a_5 x^2 &+ b_5 y^2 &+ c_5 xy &+ d_5 x & &+ f_5 \\ C_6=f_A C_B - f_B C_A=& a_6 x^2 &+ b_6 y^2 &+ c_6 xy &+ d_6 x &+ e_6 y & \\ \end{matrix}

(the parameters $(a_X, b_X, c_X, d_X, e_X, f_X)$ are the ones of the conic $C_X$)

I know nothing of them, they could be parabola, ellipses, hyperbolas, degenerates or whatever else. I need to keep everything symbolic.

Also very importantly: I am looking for a closed form.

I found this post a while ago, but even thought the original question is answered, I am still missing a lot of pieces of the puzzle.

I want to stay as much as possible in the pencil method (for simplicity and elegance's sake) so for now I am not looking for a solution using a quadratic equation (but feel free to link me some anyway if you think that's better).

Hence I want to find $\lambda \in \mathbb{R}$ so that $\det(C_1 + \lambda C_2) = 0$ .

This is a 3rd degree polynomial in $\lambda$, and I am looking for the guaranteed real root.

My immediate problem is: it is (relatively) easy to find closed form for the roots, but how can I find the real root?

Until now I have tried to exploit this paper to find the roots, but the redefinition of the cube root is very unpractical in my case.

I am also thinking that there might be a way to lower the degree of the polynomial, or to at least make it easier to solve, by applying geometric transformations such as translations, rotations and anisotropic scaling, or maybe by using more of the 6 conics.

Does anyone have any idea of how this could be done? I would greatly appreciate it if you would link me some papers that could be of any help.

Technolapin
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  • Any two non-degenerate in the pencil intersect in the same four points. Over ${\Bbb C}$ if you can find two of the six with a corresponding 3 by 3 matrix with non zero determinant do the grobner basis: generally it will be of the form (for $x>y$) $\langle p_4(y), cx+q_3(y)\rangle$ where the roots of $p_4(y)=0$ give you the $y$-coordinates of the four points. Over ${\Bbb R}$ there are no guarantees, if the coordinates above are all non-real you don't get real points. If you're lucky you have hyperbolas (the 2 by 2 upper minor negative), and not $x^2+y^2+1=0,$ or just not intersecting ellipses. – Jan-Magnus Økland Oct 31 '22 at 13:05
  • This looks interesting, I will give it a try, thank you – Technolapin Oct 31 '22 at 18:59

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