## Reflexion on conics

Given a conic (c) and a line (a) intersecting the conic at points A1, A2, consider the following transformation on the points of (c): For a point B1 on the conic take its reflexion on line (a) and define B2=H(B1) to be the intersection point of (c) with the reflected line A1B1 on (a). Show that lines B1B2 pass all through a fixed point F.

This is an immediate consequence of the fact that H is an involutive homography (see InvolutiveHomography.html ). F is the corresponding Fregier point of the homography and can be easily located. In fact, let C be the intersection point of (c) with line (d) orthogonal to A1A2 at A1. F is the intersection point of the two tangents of (c), at C and at A2. This follows by showing that besides A2, point C is also a fixed point of H, thus line CA2 is the homography axis of H, hence F is its pole.
Note that the role of C and A2 can be interchanged, giving the same Fregier point F.
Note also that fixing A1 and moving A2, the corresponding homography axis CA2 passes through a point G(A1), coinciding with the Fregier point of the involution of turning a right angle about A1 (see Fregier.html ). Thus, for A2 moving on the conic, the corresponding F draws a line e(A1), coinciding with the polar of G(A1).
Note that F is on line A1A3 which is the reflexion of the tangent a1 with respect to A1A2. This because A3 is H(A1). Thus, F is at infinity only if the angles of the tangents a1, a2 at A1, A2 respectively are equal i.e. A1, A2 are parallel to the axes.
Last remark can be used to give a simple procedure to locate the oscillating circle of a conic at a given point. This is discussed in OscillatingCircle.html .
See the file RotationOnConics.html for the generation of a homography reminiscent of the usual rotation as a product of two reflexions passing through the rotation center and intersecting there at half the rotation angle.