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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!

Arctic Char
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zooby
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    You tagged it algebraic geometry, but I presume you're working with real conics here. – Ted Shifrin Jan 18 '20 at 19:10
  • @Ted Yes indeed. – zooby Jan 18 '20 at 19:13
  • There can be up to four real intersections. In general, this involves solving a quartic equation, although there’s an algorithm that reduces it to solving a cubic. As there’s no “neat” formula for solving cubics or quadrics, there’s not going to be one for this intersection, either. – amd Jan 18 '20 at 19:24
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    This is essentially a duplicate of https://math.stackexchange.com/q/425366/265466. Although that asks specifically about ellipses, many of the methods, in particular the algorithm described in this answer apply to the intersection of any two conics. – amd Jan 18 '20 at 19:27
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    @amd The question is about determining whether there are intersections, not about determining the solutions of the equations themselves. – zooby Jan 18 '20 at 20:18
  • @amd It's a good but it does involve solving a cubic equation. – zooby Jan 18 '20 at 20:19
  • @zooby True. I missed that the OP isn’t necessarily looking for the intersection points themselves. One can use the discriminant of the resulting quadric/cubic to determine the number of its real roots, but that’s not particularly “neat,” either. – amd Jan 18 '20 at 22:02
  • See also https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem#Examples. – amd Jan 18 '20 at 22:02
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    Also, I’m not entirely convinced that the discriminant of the single-variable equation that is derived from the system necessarily tells you about the number of real intersections. Just as intermediate complex values arise when solving a cubic that has all real roots, complex roots of the resultant could still lead to real solutions of the system of conic equations. – amd Jan 18 '20 at 22:15
  • Doesn't reduce perform quantifier elimination (QE)? Have you tried other QE software, e.g, QEPCAD? Do not be too surprised if the QE software outputs a quantifier-free formula that fills 1000 pages of a PDF file. – Rodrigo de Azevedo Jan 19 '20 at 21:14
  • Rodrigo, I did see QEPCAD, and it does what I want. Except I haven't tried it because it's Linux only. Also, I think the answer would not be in a 'neat' vectorish form but like you said a very long 100 page expression! But yes, it would be like writing a tensor expression component by component instead of a neat tensor equation. Also yes you would reduce the Existence expression for the solution of real values for x and y. – zooby Jan 20 '20 at 04:25
  • I think I should read this book: https://www.maa.org/press/maa-reviews/real-solutions-to-equations-in-geometry – zooby Jan 20 '20 at 04:52
  • @zooby Try this one. – Rodrigo de Azevedo Jan 20 '20 at 13:30

2 Answers2

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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 $t$.

Jean Marie
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  • That seems more practicle. It is theoretically possible to determine the solution just with polynomials of the coefficients, but probably not numerically tractable. – zooby Jan 20 '20 at 04:30
  • I had forgotten to say that you can use tangent half angle formulas (see my Edit). – Jean Marie Jan 20 '20 at 06:39
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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.

zooby
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