When do two conic sections intersect?

Question

Given two conics

$$ax^2+bxy+cy^2+dx+ey+f=0$$

and

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

I want to find the conditions that they have a common intersection point. Unfortunately I think the answer may involve enormous polynomials of degree $>12$. (For example the Reduce function in Mathematica simply gives up!)

To make it 'simpler' I consider writing these as vector formula with $X=(1,x,y)$ then they can be written $P^{ij}X_iX_j=0$ and $Q^{ij}X_iX_j=0$. Then the solution would only depend on the pair of $3\times3$ matrices $P$ and $Q$.

I have searched everywhere and can't find a solution so I expect this is a very hard question that may even be unsolved. Is there any known solution? Particularly a 'neat' solution in terms of matrices would be nice!

Answer 1

I understand that your are asking your question in the hope that there is a closed form solution. My experience is that it is out of reach, and inefficient.

A method I have experienced in particular for easy plotting of conic curve is to switch to polar representation : $x=r \cos \theta, y=r \sin \theta$.

In this way, you get a quadratic in $r$ : $Ar^2+Br+C=0$ where $A,B$ and $C$, functions of $\theta$ in general are considered as parameters. In the case the discriminant $\Delta$ has a domain for variable $\theta$ where it is positive, you have a real conic curve possibly with two expressions $r=\tfrac{1}{2A}(-B\pm \sqrt{\Delta})$

Do this for your 2 conics. You have now to solve trigonometric equations (with an 's') which, sometimes are interesting in themselves. For example, in some cases, it is possible to bound each equation and be able to prove in this way that no intersection is possible.

In other cases, if it is possible for you to move the origin to a focus of one of the two conics, you may know that its polar equation is especially simple :

$$r=\dfrac{p}{1-e \cos(\theta-\theta_0)}$$

where $e$ is the conic's eccentricity, and $p$ is the distance focus-directrix.

Edit : Tangent half angle formulas (https://en.wikipedia.org/wiki/Tangent_half-angle_substitution) : $\cos \theta = \dfrac{1-t^2}{1+t^2}$ $\sin \theta = \dfrac{2t}{1+t^2}$ can be helpful : they permit to transform any polynomial in $\cos \theta, \sin \theta$ (as $A,B,C$ are) into a rational fraction in variable $t$.

Answer 2

I have found the answer.

Define:

$$I_{3,0} = P^{ij}P^{kl}P^{qt}\varepsilon_{ikq}\varepsilon_{jlt}$$ $$I_{2,1} = 3P^{ij}P^{kl}P^{qt}\varepsilon_{ikq}\varepsilon_{jlt}$$ $$I_{1,2} = 3Q^{ij}Q^{kl}P^{qt}\varepsilon_{ikq}\varepsilon_{jlt}$$ $$I_{0,3} = Q^{ij}Q^{kl}Q^{qt}\varepsilon_{ikq}\varepsilon_{jlt}$$

Then two ellipses in standard form, intersect when:

$$ 27 I_{03}^2 I_{30}^2 - 18 I_{03} I_{12} I_{21} I_{30}- 4I_{12}^3 I_{30}-4I_{03} I_{21}^3 - (I_{21})^2(I_{12})^2>0$$

There are some special cases and it's not very numerically stable since it is a polynomial of degree 12.