ConicThroughQuadratic

Conic through quadratic correspondence

Given two vectors a and b, consider the points on their lines {f_x = f(x)*a, g_x=g(x)*b}, where the functions are given by the simple formulas.
1) f(x) = (A*x)/(C*x + D).
2) g(x) = (A1*x² + B1*x)/(C1*x + D1).
Then the line L_x passing through these points {f_x,g_x} envelopes a conic.
Construct this conic.

The construction illustrated in the above figure is carried out as follows.
1] To show that the envelope is a conic reduce it to a Chasles-Steiner envelope for lines joining points in homographic relation.
(See Chasles_Steiner_Envelope.html ). For this show that there is a third vector c, such that its line is intercepted by L_x at points e_x, such that the correspondence f_x --> e_x is homographic.
2] This needs some calculation but is not problematic to show.
3] It is also relatively easy to show that the vector c is collinear with the limiting line L₀, obtained for x --> 0.
4] According to Chasles-Steiner the enveloping conic is tangent to the lines generated by {a, c}.
5] Further it is easy to find the point of contact f₀ of the contact with the first line (identified with the x-axis).
6] Equaly easy is determined the point g₀ for which the tangent is parallel to the x-axis.
The formulas above give hints on how this is done.
7] It is also easy to find the tangents parallel to the second line, intercepting the x-axis at f₁ and A/C respectively.
8] Then the parallelograms circumscribed and inscribed in the conic and starting at f₀ are immediately constructible.
9] The conic is determined as the member of the bitangent family determined by the triangle f₀pf₁ and passing through q.

Conic through quadratic correspondence

See Also