L(X)=mx+ny+pz=0, and conics are represented by symmetric quadratic forms q(X,X)=ax

I denote by the same letter q(X,X') the corresponding bilinear form:

q(X,X') = axx'+byy'+czz'+f(yz'+y'z)+g(zx'+z'x)+h(xy'+x'y).

The property to show is the following one [SalmonConics, p. 270]:

Let a line L(X)=0 intersect the conic q(X,X)=0 at two points {X

joining a point X

quadratic equation:

(0) L

To prove this consider an arbitrary line through X

(1) X' = kX

The intersection point of this with L(X')=0 satisfies:

(2) kL(X

If this point is also on the conic then q(X') = 0 translates to:

(3) 0 = q(X',X') = q(-L(X)X

which is the claimed equation.

This simple property some interesting applications discussed below.

legs on the conic passes through a constant point Y

To prove this use the

coordinates i.e. replacing the cartesian (x', y') by (x'=x/z, y'=y/z) (see ProjectiveCoordinates.html ). In this system

if a product of lines L(X)*L'(Y) = (mx+ny+pz)(m'x+n'y+p'z) consists of

(4) mm' + nn',

represents the inner product of the normals of the two lines, this must be zero. The inverse is also true and gives

a necessary and sufficient condition for a degenerate quadratic to represent the product of two

lines (in the natural homogeneous coordinate system).

Apply this criterion to the previous configuration, for which X

(5) -2L(X)q(X,X

is the degenerate quadratic giving the product of the lines L

easily seen to be:

(6) -2(amx

Thus the variable line L(X) = mx+ny+pz passes through the point:

(7) Y

In fact (7) shows that points Y

represented in homogeneous coordinates through the diagonal matrix diag( b-a, a-b, a+b ).

legs on the conic envelopes a second conic c'.

In this case the line joining the intersection points {x,x'} of the legs of the angle with the conic has coefficients

(r,s,t) and satisfies the equation of the first section:

(rx+sy+tz)

Again the equation mm'+nn'=0 for the sum of the coefficients of the terms x

leads this time to a quadratic equation with respect to (r,s,t) with coefficients depending on the coordinates (x

and the coefficients (a,b,c,f,g,h) of the conic. Thus the line envelopes a conic.

First the sum of the coefficients of the terms x

Equating the sum of the coefficients to zero we obtain the equation in (r,s,t):

Fregier.html

Produced with EucliDraw© |