For the practical construction of the quadratic transformation of the family generated by two conics (a) and (b) apply the following recipe:

1) For an arbitrary point P consider the two polars L

2) Define Q as the intersection point of the two lines L

To prove the claim on the existence of this transformation, work in the standard model of projective plane and consider the representation of the conics through symmetric matrices, so that conics (a), (b) are represented respectively through equations of the form x

x

Given a point [x] (projective coordinates [x] = [x

xA

In order to find a [y] satisfying (*)

x

Thus, to solve (*) it suffices to solve (**) i.e. find a z simultaneously orthogonal to the given x

F(x) = A

This is a homogeneous quadratic function in the coordinates (x

The transformation is well defined everywhere except the points x for which x

Obviously, the image c = F(L) of a line L under a quadratic transformation is a conic and introducing projective coordinates on line L we obtain, through the quadratic transformation, a rational representation of the corresponding conic c.

In the generic case, in which A

[1] The reciprocity of pole-polar implies that if Q = F(P) then P = F(Q) i.e. F is involutive, its inverse coinciding with itself.

[2] By Desargues involution theorem, the conics of the family I(a,b) define on line PQ (with Q = F(P)) through their intersection points an involution. P, Q are the fixed points of this involution, since they are conjugate with respect to every member of the family. Consequently the members of the family passing through P, Q are tangent there to the line PQ.

For a continuation of this subject look at the file Quadratic_Transformation2.html .

DesarguesInvolution.html

ElevenPointConic.html

Polar.html

Quadratic_Transformation2.html

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