Chasles_Steiner

1. Chasles-Steiner definition of a conic

The Chasles-Steiner definition of a conic ([ChaslesConiques, p.5], [Baker, p. 73]), illustrated by the figure below, can be described as follows.
Consider the two pencils A* and B* of lines through points A and B correspondingly and a homographic relation L'=F(L) between the lines of the two pencils.
The Chasles-Steiner method of definition describes a conic as the geometric locus of intersection-points P of lines L and L'.

2. Pencil parameterization

The pencil A* of all lines through point A can be parameterized by the points of an arbitrary but fixed (projective) line E. To every line L through Acorresponds the intersection point Q of L with E. Introducing (projective) coordinates for the line E we can correspond to L a definite number x. Various such coordinate systems, corresponding to different positions of the parameterizing line E, are related by homographic relations x'=(ax+b)/(cx+d) (see HomographicRelation.html ).

3. A proof of the claim

To prove the claim of section-1, on the generation of conics, I use a projective coordinate system, namely barycentric coordinates (see BarycentricCoordinates.html ) with respect to a triangle with vertices {A,B,C}. Point C is taken to be one of the points of the locus i.e one of the intersection points of line L with line F(L). The pencils are parameterized with the aid of a line E passing through C.

In such a coordinate system points of the line E are represented as C+tD and the homographic relation is given by a function f(t)=(at+b)/(ct+d).
Since C is mapped to itself f(0)=0 i.e. b=0. D is a fixed point on line E. Writing D in the form D=pA+qB+rC we have for X the projective coordinates (tp,tq,1+tr) and for Y=F(X) the coordinates (f(t)p,f(t)q,1+f(t)r). Line L has coefficients given by the vector product (1,0,0)x(tp,tq,1+tr)=(0,-1-tr,tq).
Line L' has coefficients given by the vector product (0,1,0)x(f(t)p,f(t)q,1+f(t)r)=(1+f(t)r,0,-f(t)p).
Their intersection point is given by the vector product (0,-1-tr,tq) x (1+f(t)r,0,-f(t)p) = ((1+tr)f(t)p, tq(1+f(t)r), (1+tr)(1+f(t)r)).
Equating this vector with (kx,ky,kz) and eliminating {t,k} we find after a short calculation
                                                                                              (c+r(a-d))xy + (-ap)yz + (dq)zx =0.
This is the equation of a conic passing through the three base points {A,B,C} of the coordinate system. See the file ChaslesSteinerExample.html for another example where a similar calculation is carried out giving a characteristic property of hyperbolas. See also the file Maclaurin_Chasles_Steiner.html for anexploration of the relation of this theorem to the earlier one by Maclaurin (see Maclaurin.html ).

4. The degenerate case

Without loss of generality we may assume that point D coincides with the intersection of the parameterizing line E with line AB. From the expressionD = pA+qB+rC, this is equivalent with the condition r = 0 and implies the simpler form of the equation of the conic
                                                                                                 (c)xy + (-ap)yz + (dq)zx = 0.
The determinant of the corresponding matrix representing the conic in these coordinates (up to the factor -2) is  Det = cadpq.
The conic is degenerate (product of two lines) if and only if Det = 0, thus giving the condition  cadpq = 0. Here {p,q} are different from zero, since otherwiseD would coincide with A or B. Hence the condition is equivalent with  cad = 0.  Again {a,d} are different from zero, since otherwise the homographic relation would be a constant one. Thus Det = 0 is equivalent with c = 0 and the matrix equation reduces to
                                                                                                 z(dqx - apy) = 0.
which expresses indeed the product of the two lines
                                                                                                 z = 0  and   (dq)x - (ap)y = 0.
The first line coincides with AB and the second passes through C and the point on the line AB:  D' = (ap)A + (dq)B. Thus we come at the result
If the Chasles-Steiner generation produces a degenerate conic then one of the lines is the line through the poles (A,B) of the procedure.

Remark A criterion of degeneration is the discovery of three collinear points lying on the locus described by the Chasles-Steiner procedure. This remark has many applications to particular configurations. An example is given in the next section. See also Chasles_Steiner_Line_Locus.html .

5. Line locus

Given a triangle ABC a line p and two points {B', C'} on side BC consider all points P on line p. Let {B_x, C_y} denote the intersection points of lines {B'P, C'P} with sides{AB, AC} respectively. Assume that {x,y} denote the coordinates of these points for systems of line coordinates along {BA, CA} starting respectively at {B,C}. Then these coordinates satisfy a relation of the form y = ax/(bx+c).
The inverse is also valid, i.e. if on sides {BA, CA} points {B_x, C_y} are taken, in such a way that their coordinates {x,y} satisfy the above relation, then lines {B'B_x, C'C_y}intersect at a point P describing a line p, as x varies.

The proof of the property follows by composing two homographic relations describing the perspectivities through points {B',C'}. The first corresponds to B_x of line AB the point point P on line p. The second corresponds to P on lin p point C_y. The composition of the two is the correspondence B_x --> B_y described by the composition of two Moebius transformations which is also a Moebius transformation, thereby proving the claim (see HomographicRelation.html , HomographicRelationExample.html ).
The inverse follows from the previous section. According to it if {B_x, C_y} are related by such a relation, then the intersection point of lines {B'B_x, C'C_y} describes aconic which passes through the points {B',C'}. But the conic passes also through the limit point F on BC resulting for x-->0 (can be computed analytically see calculation below). Hence the conic contains a line (BC) and must be degenerate, thus, P describes a second line passing through F.

To find the limit position F of P for x-->0 project P on BC to F' (not drawn). Then PB' = PF'/sin(B'), PC' = PF'/sin(C'). From triangle BB'B_x it is(sine-theorem) (x/sin(B')) = BB'/sin(B_x) => sin(B') = (x/BB')sin(B_x). Analogous expression for sin(C') gives PB'/PC' = sin(C')/sin(B') =((y/CC')sin(C_x))/((x/BB')sin(B_x)) and since the limit y/x for x-->0 is (a/c) the above equality gives that FB'/FC' = (a/c)(BB'/CC')(sin(C)/sin(B)).

Bibliography

[Baker] Baker, H. F. Plane Geometry New York, Chelsea Publishing Company 1971
[ChaslesConiques] Chasles, M. Traite des sections coniques Gauthier-Villars, Paris, 1865

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