[1] Draw OP' and the chord QQ' through P and parallel to OP'. Q' is a fixed point on the circle.

[2] OAP'P and OP'PQ' are parallelograms. Quadrangle OPQP' is an isosceles trapezium.

[3] The intersection point W of the diagonals of the trapezium describes a parabola

[4] This parabola is the

The proofs are easy.

[1] follows from the fact that drawing OQ' orthogonal to a, the resulting OP'PQ' is a parallelogram.

[2] follows from [1] and the equality OQ = OA = P'P = r.

[3] follows from the fact that the medial line t

[4] follows from the fact that t

(i) F fixes points {B,C, O} and

(ii) F maps A to W

- It follows easily (exercise) that F fixes every point of line BC and also leaves invariant every line passing through O (maps such a line into itself fixing O and its intersection point with BC).

- Using this and the invariance of line AO one sees easily that every line parallel to BC maps to a line also parallel to BC. In particular the parallel to BC through A maps to the parallel to BC through W

- Using cross ratios allong line AO one sees easily that the tangent line t of the circle at Q' maps via F to the line at infinity.

- From the previous result follows also immediately that the tangents {t

- From these facts follows, that points {A,B,C} and the tangents to the circle at these points map to {W

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