Given four points in the plane, there exists a one-dimensional family of conics through these, often called a pencil of conics. The locus of the centers of symmetry for all of these conics is again a conic. What's the most elegant way of computing it?

I know I could choose five arbitrary elements from the pencil, compute their centers and then take the conic defined by these. I can also do so on a symbolic level, to obtain a general formula. But that formula is at the coordinate level, and my CAS is still struggeling with the size of the polynomials involved here. There has to be a better way.

Bonus points if you know a name for this conic. Or – as the center is the pole of the line at infinity – a name for the more general locus of the pole of an arbitrary line with respect to a given pencil of conics.

  • 1
    $\begingroup$ Your question is very interesting. I will think about it. Is this site interesting for your issue ? $\endgroup$ – Jean Marie Nov 10 '16 at 23:39
  • $\begingroup$ @JeanMarie: so far I don't see anything on that site which might help me here, although it does cover a number of key concepts I was already aware of. In the meantime I realized that I should be able to apply the same technique I used to describe a different conic, so I'm currently in the process of streamlining that computation to make it easier to apply to new situations. $\endgroup$ – MvG Nov 13 '16 at 10:39
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    $\begingroup$ "In 1912 Maud Minthorn showed that the nine-point conic is the locus of the center of a conic through four given points." in (en.wikipedia.org/wiki/Nine-point_conic) $\endgroup$ – Jean Marie Nov 13 '16 at 11:41
  • 1
    $\begingroup$ @Narasimham: For every point $P_5$ in the plane, the points $P_1$ through $P_5$ will define a unique conic, and it's the locus of all of these I'm after. So the locus of $P_5$ would be best described as the whole plane, excepting degenerate situations like $P_5$ coinciding with one of the other defining points, or four of these points being collinear. Restricting $P_5$ to some other locus may restrict the resulting centers to only part of the conic I'm after. $\endgroup$ – MvG Nov 13 '16 at 21:48
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    $\begingroup$ Given four points $P_1,P_2, P_3,P_4$ in general position. If you take $P_5 = P_i + P_j - P_k$ where $i \ne j \ne k \in \{ 1,2,3,4 \}$, the center of the conic passing through $P_1,\ldots,P_5$ will be $\frac12 ( P_i + P_j )$. If you know the locus of center is a conic, this gives you $6$ easy to compute points for constructing it. $\endgroup$ – achille hui Nov 14 '16 at 19:34

This is a rough sketch, there might be some imprecision.

What about something like this. Take, let's say, the three by three matrix $A$ representing one degenerate conic from the pencil and let $B$ be another three by three matrix representing another degenerate conic from the pencil (you can form them from the appropriate pairs of points among the four points that determines the pencil). Then the pencil of conics is defined by the projective family of three by three matrices $C_{[\lambda : \mu]} = \lambda \, A + \mu \, B$ so that a conic is given by the equation $x^T \big(C_{[\lambda : \mu]}\big) x = 0$ with $x \in \mathbb{C}^3 \setminus \{0\}$. Now, your line at infinity is given by a (dual) vector $l \in \mathbb{C}^3 \setminus \{0\}$ in the form $l^T x = 0$. The pole $P(\lambda,\mu)$ of the line $l$ with respect to the conic $C_{[\lambda : \mu]}$ is given by $$P(\lambda, \mu) = \big(C_{[\lambda : \mu]}\big)^{-1}\, l$$ However, $$\big(C_{[\lambda : \mu]}\big)^{-1} = \frac{1}{\det\big(C_{[\lambda : \mu]}\big)} \,\, \text{Adj}\big(C_{[\lambda : \mu]}\big)$$ where $\text{Adj}\big(C_{[\lambda : \mu]}\big)$ is the adjoint matrix of $C_{[\lambda : \mu]}$ i.e. $$\text{Adj}\big(C_{[\lambda : \mu]}\big) \, \big(C_{[\lambda : \mu]}\big) = \big(C_{[\lambda : \mu]}\big) \, \text{Adj}\big(C_{[\lambda : \mu]}\big) = \det\big(C_{[\lambda : \mu]}\big)\, I$$ The entries of $\text{Adj}\big(C_{[\lambda : \mu]}\big)$ are the two by two minors of $C_{[\lambda : \mu]}$, with appropriate signs, and so quadratic homogeneous polynomials with respect to $(\lambda,\mu)$. Since we work in the projective plane $\mathbb{CP}^2$, multiplying the pole $P(\lambda,\mu)$ by $\det\big(C_{[\lambda : \mu]}\big)$ gives us the same projective point, so we can actually write the pole as $$P(\lambda,\mu) = \text{Adj}\big(C_{[\lambda : \mu]}\big) \, l$$ Now this is exactly a $(\lambda,\mu)$ parametrization of the conic of poles of $l$ with respect to the pencil of conics.

If you want it written as equations, rather than parametrized, then the conic is written as parametrized equations \begin{align} x_1 &= P_1(\lambda,\mu)\\ x_2 &= P_2(\lambda,\mu)\\ x_3 &= P_3(\lambda,\mu) \end{align} where $P_j(\lambda,\mu)$ are homogeneous polynomials of degree two. Write everything in the affine chart \begin{align} \frac{x_1}{x_3} &= \frac{P_1(\lambda,\mu)}{P_3(\lambda,\mu)}\\ \frac{x_2}{x_3} &= \frac{P_2(\lambda,\mu)}{P_3(\lambda,\mu)} \end{align} So if $s = \frac{\lambda}{\mu}$ then we can write everything as \begin{align} x = \frac{x_1}{x_3} &= \frac{P_1(s,1)}{P_3(s,1)}\\ y = \frac{x_2}{x_3} &= \frac{P_2(s,1)}{P_3(s,1)} \end{align} or alternatively \begin{align} P_1(s,1) - P_3(s,1) \, x &= 0\\ P_2(s,1) - P_3(s,1) \, y & = 0 \end{align} where again, $P_j(s,1)$ are polynomial of degree two in $s$. Now, write the resolvent equal to zero of the polynomials $P_1(s,1) - P_3(s,1) x $ and $P_2(s,1) - P_3(s,1) y$ viewed as polynomials of one variable $s$, thus eliminating $s$, and you get an equation for $(x,y)$.

I hope it makes sense.

  • $\begingroup$ This is a lot like the coordinate-based approach I had in mind, except that instead of eliminating a parameter, I found that choosing five explicit elements of the pencil and computing the conic defined by their centers was faster in my symbolic computation. $\endgroup$ – MvG Nov 13 '16 at 10:36
  • $\begingroup$ I praise this interesting solution because it is only based on projective concepts. For less initiates, they should first have a look at (en.wikipedia.org/wiki/…) with in particular the fact that the center of a conic C is the pole of the line at infinity. $\endgroup$ – Jean Marie Nov 13 '16 at 11:22
  • $\begingroup$ See the reference I have given to the OP. $\endgroup$ – Jean Marie Nov 13 '16 at 14:57
  • $\begingroup$ @JeanMarie I actually discovered this fact when I was thinking about the problem. It is not that surprising. In projective terms, you take the 9 points to form harmonic cross-ratios (cross-ratios -1) with any pair $M,N$ of the four points and the point of intersection of line $l$ with line $MN$. This just follows from the way the polar of a polar line is determined in a degenerate conic (the polar is not a point, it is a line). However, I thought that the OP didn't want that, because it leads again to fitting a conic through five out of the nine points... $\endgroup$ – Futurologist Nov 13 '16 at 15:22

While there were many good answers here, I feel that two comments were even more useful to me, so I'll combine them to a CW answer.

@JeanMarie quoted Wikipedia:

In 1912 Maud Minthorn showed that the nine-point conic is the locus of the center of a conic through four given points.

So this gives a name to the conic I described. It also characterizes it as a conic passing through the centers of the sides as well as through the points of intersections of opposite sides.

The former of these properties is something @achille hui deduced from my specification of the conic:

Given four points $P_1,P_2,P_3,P_4$ in general position. If you take $P_5=P_i+P_j-P_k$ where $i\neq j\neq k\in\{1,2,3,4\}$, the center of the conic passing through $P_1,\dots,P_5$ will be $\frac12(P_i+P_j)$. If you know the locus of center is a conic, this gives you $6$ easy to compute points for constructing it.

So yes, taking the midpoint for every pair of points, dropping one of these and then constructing the conic through the other five is an elegant approach.

It is not perfect in terms of simplicity of the resulting formulation, though. Labeling the points $A,B,C,D$ and taking all pairwise midpoints except for $\frac12(C+D)$ (or rather $D_z\cdot C + C_z\cdot D$ in homogeneous coordinates), I found the following removable common factor in a homogeneous formulation of this construction:

$$A_z^3\cdot B_z^3\cdot C_z\cdot D_z\cdot\begin{vmatrix} A_zB_x-B_zA_x & C_zD_x-D_zC_x \\ A_zB_y-B_zA_y & C_zD_y-D_zC_y \end{vmatrix}$$

The first four factors indicate that the construction outlined above has a removable singularity (resulting in a null matrix instead of some conic) if one of the defining points is at infinity. This should not be a problem in non-projective scenarios where all inputs are guaranteed to be finite. It's also not surprising as the midpoint of a segment with one endpoint at infinity will be said point at infinity, so three of the six pairwise midpoints will be identical.

The last term, the determinant, characterizes a situation where the line $AB$ and the line $CD$ are parallel. In that case, four of the five defining points would lie on the parallel halfway between $AB$ and $CD$, and the remaining midpoint of $AB$ is not enough to define the second line of this degenerate conic. So be sure that the midpoint you omit corresponds to a line which is not parallel to its counterpart.

My other answer discusses a formulation which avoids these removable singularities.


Here's a coordinate derivation that's computationally intense, although my copy of Mathematica was able to crunch the symbols fairly quickly.

Let the points $P$, $Q$, $R$, $S$ be arranged such that segments $\overline{PR}$ and $\overline{QS}$ meet at the origin, and such that the coordinate axes bisect the angles created by those segments. (This will give us a nice symbolic symmetry in $x$ and $y$.) Thus, we may assign $$\begin{gather} P = p\;(\phantom{-}\cos\theta, \phantom{-}\sin\theta) \quad Q = q\;(\phantom{-}\cos\theta, -\sin\theta) \\ R = r\;(-\cos\theta, -\sin\theta) \quad S = s\;(-\cos\theta,\phantom{-}\sin\theta )\end{gather}$$

(So, for positive $p$, $q$, $r$, $s$, we have opposite rays $\overrightarrow{OP}$ and $\overrightarrow{OR}$, and opposite rays $\overrightarrow{OQ}$ and $\overrightarrow{OS}$, emanating from the origin.)

We'll parameterize the one-parameter pencil of conics by conveniently parameterizing the variable fifth point that defines the members of the pencil: $$T := t\; ( P + Q )$$

Mathematica tells me that the corresponding conic's equation ... $$a x^2 + b x y + c y^2 + d x + e y + f = 0$$ ... has these coefficients: $$\begin{align} a &= -\phantom{2}(r s + (q r + p s - 2 r s) t + (p - q )(r-s) t^2) \sin^2\theta \\ b &= \phantom{-} 2 (p r - q s) t^2 \sin\theta \cos\theta\\ c &= \phantom{-2}(r s + (q r + p s - 2 r s) t - ( p + q )( r + s ) t^2) \cos^2\theta \\ d &= \phantom{-} 2 (p q r + p q s - p r s - q r s) t^2 \sin^2\theta \cos\theta\\ e &= - 2 (p q r - p q s - p r s + q r s) t^2 \sin\theta\cos^2\theta \\ f &= \phantom{-} 4 p q r s t^2 \sin^2\theta \cos^2\theta \end{align}$$

Now, referring to this tremendously-convenient answer, we have that the conic's center has coordinates

$$x = \frac{b e - 2 c d}{4 a c - b^2} \qquad y = \frac{b d - 2 a e}{4 a c - b^2}$$

The numerators and denominators of these expressions are slightly-hairy fourth-degree polynomials in $t$ that I won't transcribe. From here, I invoked Mathematica's Resultant[] function to eliminate the parameter $t$; the result(ant), after ignoring extraneous factors, is:

$$\begin{align} 0 \quad=\quad &\phantom{2} x^2 ( p r - q s )\sin^2\theta \\ +\; &\phantom{2} y^2 ( p r - q s ) \cos^2\theta \\ -\; &2 x y ( p r + q s ) \sin\theta \cos\theta \\ +\; &\phantom{2} x ( (p-r) qs - pr(q-s) ) \sin^2\theta \cos\theta \\ +\; &\phantom{2} y ( (p-r) qs + pr(q-s) ) \sin\theta \cos^2\theta \end{align} \tag{$\star$}$$

With no constant term, $(\star)$ describes a conic through the origin; that is, the conic-of-centers passes through the intersection of $\overline{PR}$ and $\overline{QS}$. Of course, if the conic passes through the intersection of one pair of segments determined by the original points, then it must pass through the intersections of the other two pairs of segments, as well. (Symbolic verification is left to the reader.) Thus, three points of the conic-of-centers are easily determined geometrically ... and they might even be "obvious" in a projective approach.

The all-important discriminant has the remarkably-simple form:

$$16 p q r s \sin^2\theta \cos^2\theta$$

from which we can say

$$\text{the conic-of-centers is} \begin{cases} \text{an ellipse} &\text{if}\;pqrs < 0 \\ \text{a parabola} &\text{if}\;pqrs = 0 \\ \text{an hyperbola} &\text{if}\;pqrs > 0 \end{cases} $$

The eccentricity computation is messy, so I'll leave that to the reader. On the other hand, the center is quite nice: it is, in fact, the centroid of $P$, $Q$, $R$, $S$:

$$\text{center of the conic-of-centers}\; = \frac{1}{4}( P + Q + R + S )$$

I guess I'll stop here.

One more thing ... If $p$, $q$, $r$, $s$ are all non-zero, then we can define reciprocals $\overline{p} := 1/p$, etc, and divide $(\star)$ through by $pqrs$ to get this form of the equation:

$$\begin{align} 0 \quad&=\phantom{2} x^2 ( \overline{p}\overline{r} - \overline{q}\overline{s} )\sin^2\theta \\ &+\phantom{2} y^2 ( \overline{p}\overline{r} - \overline{q}\overline{s} ) \cos^2\theta \\ &+2 x y ( \overline{p}\overline{r} + \overline{q}\overline{s} ) \sin\theta \cos\theta \\ &+ \phantom{2} x ( (\overline{p}-\overline{r}) - (\overline{q} - \overline{s}) ) \sin^2\theta \cos\theta \\ &+ \phantom{2} y ( (\overline{p}-\overline{r}) + (\overline{q} - \overline{s} )) \sin\theta \cos^2\theta \end{align} \tag{$\overline{\star}$}$$

  • $\begingroup$ See the reference I have given to the OP. $\endgroup$ – Jean Marie Nov 13 '16 at 14:58
  • $\begingroup$ @JeanMarie: I saw it. Good catch. It seems I'm 100 years late and 6 points short in my analysis. :) I'm intending to update my answer to show that the six midpoints lie on the conic. (This is reasonably easy to check, since the midpoints have the form $(m\cos\theta, n \sin\theta)$, which leads to some simplification.) $\endgroup$ – Blue Nov 13 '16 at 15:13

Based on the fundamental theorem of projective geometry, it must be possible to take any projectively invariant property which can be described as a polynomial in the coordinates of the involved points and turn that into an expression in determinants of these homogeneous coordinates of these points instead. Using randomized evaluation of polynomial expressions, one can find valid combinations of relevant determinants and similar terms. I've used this technique before while formulating this question of mine, and now remembered to do it here as well.

The linear dependencies identified by my randomized evaluation look like this:

goal                    1                                                            
[B,C,D] [A,C,D] (l,AB)     1                                                         
[B,C,D] [A,B,D] (l,AC)        1                                                      
[B,C,D] [A,B,C] (l,AD)           1                                                   
[A,C,D] [A,B,D] (l,BC)              1                                                
[A,C,D] [A,B,C] (l,BD)                 1                                             
[A,B,D] [A,B,C] (l,CD)                    1                                          
<l,D> [B,C,D] (AB,AC)                        1                                       
<l,C> [B,C,D] (AB,AD)                           1                                    
<l,D> [A,C,D] (AB,BC)                              1                                 
<l,C> [A,C,D] (AB,BD)                                 1                              
<l,D> [A,B,C] (AB,CD)                                    1                           
<l,C> [A,B,D] (AB,CD)                                       1                        
<l,B> [A,C,D] (AB,CD)                                          1                     
<l,A> [B,C,D] (AB,CD)                                             1                  
<l,B> [B,C,D] (AC,AD)                                                1               
<l,D> [A,B,D] (AC,BC)  -1 -1 -1    -1       -1    -1                                 
<l,D> [A,B,C] (AC,BD)                                                   1            
<l,C> [A,B,D] (AC,BD)      1  1       -1 -1  1       -1 -1 -1          -1            
<l,B> [A,C,D] (AC,BD)                                                      1         
<l,A> [B,C,D] (AC,BD)                                                         1      
<l,B> [A,B,D] (AC,CD)     -1 -1        1  1 -1           1    -1        1 -1         
<l,D> [A,B,C] (AD,BC)                                                            1   
<l,C> [A,B,D] (AD,BC)      1    -1  1     1    -1  1     1  1                   -1   
<l,B> [A,C,D] (AD,BC)  -1 -1     1     1    -1    -1                -1  1 -1     1   
<l,A> [B,C,D] (AD,BC)                                                               1
<l,C> [A,B,C] (AD,BD)   1 -1     1     1        1     1                              
<l,B> [A,B,C] (AD,CD)   1  1    -1    -1     1          -1     1     1 -1  1         
<l,A> [A,C,D] (BC,BD)   1  1    -1    -1     1     1                   -1    -1 -1 -1
<l,A> [A,B,D] (BC,CD)     -1     1 -1    -1       -1    -1       -1              1  1
<l,A> [A,B,C] (BD,CD)                       -1           1        1     1     1      

The first column contains a non-zero entry in the row titled goal, so ignoring that entry, the rest is a formula for the conic I'm after. The other columns are just fancy descriptions of the zero matrix. Adding any of these to the first column may lead to a simpler formula in some way.

Here is how to read the notation of the terms. The letters A, B, C and D are the four points defining the pencil of conics. The letter l gives the line at infinity, which is required to make this a projectively invariant concept. The square brackets like [A,B,C] are determinants. Angle brackets like <l,D> are scalar products. If homogeneous coordinates are formed using a one in the last coordinate, and the line at infinity is described as $[0:0:1]$, then these will be just one. (As the line at infinity might be spanned by two points, this is essentially like a three-point determinant as well.) Parentheses like (AB,CD) are matrices of degenerate conics. $A\times B$ is one line joining two of the given points, and $C\times D$ is the other. So $(A\times B)(C\times D)^T$ will be a rank one matrix describing this pair of lines. Adding its transppose makes the matrix symmetric without changing the quadratic form. (l,CD) is the same thing with the line at infinity taking the role of one of the lines connecting two points. (Since the matrix constitutes a quadratic form, plugging a point X into said form would turn (AB,CD) into [ABX][CDX], i.e. again a product of determinants.)

While this makes for a shorter formula than doing this at the coordinate level, the approach does convey little geometric intuition. But perhaps playing around with the space of possible formulas will allow someone to come up with a more elegant formula which can be interpreted geometrically again.


Excuse me, I am unable to understand , since fifth point fixes the conic and its center, what is the variable fifth point $P_5$ to determine a locus of corresponding conic centers?

To find a locus we need to define our object property independently, but that property should not again depend on the what is fixed already. Kindly clarify.

For discussion reference only to a simple pencil with a fixed center at origin,

taken the coordinates passing passing through $ ( ( \pm 1,\pm 1) , (0,p) )$ in standard form

$$ a x^2 + 2 h x y + b y^2 + 2 f x + 2 g y=1. $$

Solving as a standard co-ordinate approach due to high symmetry it calculates to non-zero $(a,b)$.

$$ (p^2-1)x^2 + y^2 = p^2$$

We have the pencil sketched.


And in one more example for obtaining all conics' common center at origin with a new choice of $P_5$ set on a circle radius $2,( x = 2 \cos t, y= 2 \sin t), x= \pm 1,y= \pm 1$ we have the central conics (red) added in the sketch below.

Conics Pencil

  • $\begingroup$ Interesting. The origin is the common center for all of your pencil's conics ... except two. The centers of the "degenerate" ellipses (the pair of horizontal lines, or the pair of vertical lines) arguably have "indeterminate" centers. However, if we take the broader view that a "center" is a point that coincides with the midpoint of each chord through it, then the "degenerate" ellipses have infinitely-many centers: the points on their mid-lines. With your pencil, these are the coordinate axes. This situation is accommodated in my $(\star)$, which reduces to $xy=0$ when $p=q=r=s$. $\endgroup$ – Blue Nov 13 '16 at 16:06
  • $\begingroup$ Whoops. I meant to write "'degenerate' conics" where I wrote "'degenerate' ellipses". Of course, "'degenerate' parabolas" would be the most-appropriate term. $\endgroup$ – Blue Nov 13 '16 at 16:53
  • $\begingroup$ The max $y$ should be $p$ and that tallies everywhere. Also $p=0$ gives pair of straight lines $ y\pm x = 0,$ not sketched. $\endgroup$ – Narasimham Nov 13 '16 at 17:20
  • $\begingroup$ @Blue only 4 points are needed to determine a parabola as against 5 points for other conics. So a parabola does not/need not figure in this set at all, even as a degenerate. $\endgroup$ – Narasimham Nov 13 '16 at 17:37
  • $\begingroup$ The nine-point conic for the unit square you used should be $xy=0$. All the non-degenerate conics in the pencil have the origin as their only center of symmetry, but the pairs of parallels are symmetric with respect to any point on one of the coordinate axes, which this locus captures. And I'm with @Blue: a conic with a double point at infinity (as opposed to distinct real or complex points) would best be called a parabola, so a pair of parallel lines would be a degenerate parabola. $\endgroup$ – MvG Nov 13 '16 at 22:31

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