[alogo] Conic through quadratic correspondence

Given two vectors a and b, consider the points on their lines {fx = f(x)*a, gx=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 Lx passing through these points {fx,gx} envelopes a conic.
Construct this conic.

[0_0] [0_1] [0_2] [0_3] [0_4] [0_5]
[1_0] [1_1] [1_2] [1_3] [1_4] [1_5]
[2_0] [2_1] [2_2] [2_3] [2_4] [2_5]

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 Lx at points ex, such that the correspondence fx --> ex 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.

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