Consider the two pencils A* and B* of lines through points A and B correspondingly and a

The Chasles-Steiner method of definition describes a conic as the geometric locus of intersection-points P of lines L and L'.

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 ).

(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.

The inverse is also valid, i.e. if on sides {BA, CA} points {B

The proof of the property follows by composing two homographic relations describing the perspectivities through points {B',C'}. The first corresponds to B

The inverse follows from the previous section. According to it if {B

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

BarycentricCoordinates.html

ChaslesSteinerExample.html

Maclaurin_Chasles_Steiner.html

Maclaurin.html

Chasles_Steiner_Line_Locus.html

HomographicRelationExample.html

[ChaslesConiques] Chasles, M.

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