In the real projective plane, I am given three points and two lines. I want to find out how many conic sections there are that are incident to each of the three points and tangent to each of the two lines. What is the easiest description for how many such conic sections exist, as a function of the points and lines?

I am interested primarily in an answer for a general arrangement, that is, a solution that need not work if the points and lines are in degenerate special arrangements satisfying certain algebraic equations. I do not wish to tell what exactly these degenerate cases are, because I presume they can derived from the answer. I suspect that in a general arrangement, there are zero, two, or four such conics.

What sort of answer I'd like

To explain my motivation and what sort of answer I would like, let me bring up two analogous but simpler problem.

Suppose that I am are given four points and one line in the real projective plane, and want to find conic sections incident to each of the four points and tangent to the line.

Let's ignore the following special cases: two of the points coincide, the four points are collinear, two of the points are incident to the line, or three points are collinear and the fourth is incident to the line.

The easiest way I know to see how many such conic sections are is to transform the plane to an affine plane with a homography such that the given line becomes the ideal line. The conic section is then necessarily either a parabola or the union of two parallel lines. Four classes of arrangements are then possible.

  • If the four points form the vertexes of a convex quadrilateral in the affine plane, then there are exactly two possible conics, and both are parabolas.

  • If one of the four points is inside the triangle spanned by the other three points in the affine plane, then there is no matching conic section.

  • If three of the points are collinear and the fourth is not an ideal point, then there is exactly one solution, which is the union of the line joining the first three points and a parallel line on the fourth point. (One of the first three points may be ideal.)

  • Otherwise, one of the points is ideal and no three points are collinear, in which case there is exactly one solution, a parabola.

The kind of answer I'm looking for is a description of the possible cases for three points and two lines, plus a reference or hint for how to prove them. The last two of the above four cases are degenerate special cases. As I explained above, I am primarily looking for an answer for general arrangements, like in the first two cases above.

Why I suspect there are generally zero, two, or four solutions?

Consider the conics parametrized by the homogeneous real vector $ \mathbf{a} = (a_0, \dots, a_5) $ such that the point $ \mathbf{x} = (x, y, z)^{\mathrm{T}} $ is on the conic iff $ 0 = \mathbf{x}^{\mathrm{T}} \mathbf{A} \mathbf{x} $, where $$ \mathbf{A} = \begin{pmatrix} a_0 & a_1 & a_2 \cr a_1 & a_3 & a_4 \cr a_2 & a_4 & a_5 \end{pmatrix} $$ Notice that for any given point $ \mathbf{x} $, this condition is a homogenous linear equation over the vector $ \mathbf{a} $. For any five points in the plane, the set of conics incident to each of those are exactly the ones satisfying five such equations. It follows immediately that there must be either exactly one or infinitely many conics matching. (This is well-known.)

Now if we have the line with homogenous coordinates $ \mathbf{u} = (u, v, w)^{\mathrm{T}} $, then this is tangent to the conic iff $ 0 = \mathbf{u}^{\mathrm{T}} \mathbf{A}^{-1} \mathbf{u} $. This condition is equivalent to $ 0 = \mathbf{u}^{\mathrm{T}} (\det(\mathbf{A}) \cdot \mathbf{A}^{-1}) \mathbf{u} $. But from Cramer's theorem it follows that each of the coefficients of the matrix $ \det(\mathbf{A}) \cdot \mathbf{A}^{-1} $ are homogenous quadratic polynomials in the coefficients of $ \mathbf{a} $. As a result, the constraint that the conic is tangent to a given line is a exactly a homogenous quadratic polynomial equation on the coefficients of $ \mathbf{a} $.

Thus, if we are given one line and four points in the plane, and we want to find all conics tangent to the line and incident to the four points, then the parameters $ \mathrm{a} $ of the conic are constrained by four homogeneous linear equations and a homogeneous quadratic equation. Such an equation system must have either infinitely many solutions, or up to two solutions (up to a constant factor). Thus, either there are infinitely many matching conics, or at most two.

Similarly, in the original problem with two lines and three points, the parameters have three linear and two quadratic constraints. This means there must be either infinitely many matching conics, or at most four.

If the coefficients of the constraints were general, then we would know that there are zero, two, or four solutions of the equation system. Alas, the constraints aren't quite general, because their coefficients have nontrivial algebraic dependences from the way we created them from the points and lines. Thus, this simple argument can't prove that the constraints can't behave in some unfortunate way. It's possible at this point that there are always infinitely many or zero solutions, or that there are one or three solutions in some general arrangements, or that there are never exactly four or never exactly two solutions except in singular arrangements. In fact, you find such unfortunate behavior if you tried to apply the same argument to the case of three lines and two points: the naive argument would say that there can be up to eight solutions, but in reality there can't be more than four, not even over the complexes.


1 Answer 1


In general there will be four solutions, possibly complex. In Conic by three points and two tangent lines I've asked for ways to compute these, and based on a comment there found a way which is in my opinion pretty intuitive to understand since it relates to a 3d setup you can easily visualize. You can also use this to count possible solutions.

So when will the solutions be complex not real? Well, consider the two lines dividing the plane into two regions. In an affine setup that would be four regions, but they connect through infinity so there are only two. If all of the points are in the same region, you get four real solutions. If they are in different regions, you only get complex solutions.

If your elements are not in general position, there are further setups to consider. If the three points are collinear and within one region, then you have only a single conic (since all the planes in the idea behind the computation will coincide). If one point coincides with one of the lines, you only get two solutions. If two points coincide with the two lines you only get one solution. And I don't make any claims to completeness here at this point.

You can use this interactive web widget to experiment with this setup. It doesn't do all the degenerate cases yet, but the behavior in a small neighborhood around such a singular situation should give you a fair idea of what to expect in the limit.

  • $\begingroup$ Thank you, this answer is helpful, but I don't think it's complete. I understand that when the two lines separate any two of the three points, then there are no real solutions. But when the lines don't separate the points, will there always be four real solutions in the general case? If so, can you give me a pointer for how to prove that? $\endgroup$ Sep 19, 2016 at 8:35
  • $\begingroup$ If you follow the answer from my linked post, it boils down to the fact that three points in the plane will lift to three pairs of points on a cone, from which one can form $2^3$ planes. Two planes lead to distinct conics if projected back to the plane, unless they are symmetric wrt. the drawing plane. But they come in symmetric pairs, resulting in the $2^3/2=4$ distinct conics in general. I could give more details for each of these steps, but I don't see any of them as particularly problematic. $\endgroup$
    – MvG
    Sep 19, 2016 at 8:48
  • $\begingroup$ Thanks, the method in that post indeed makes it clear that all four solutions are real in that case. I suggested an edit to the post to state that explicitly. $\endgroup$ Sep 19, 2016 at 8:56
  • $\begingroup$ So there's always 0 or 4, never 2? That's interesting! Reminds me of arxiv.org/abs/1211.7160 . $\endgroup$ Sep 19, 2016 at 11:59

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