ReflexionOnConics

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.

Reflexion on conics

See Also