## Rectangular hyperbola member

To construct the rectangular hyperbola member of the bitangent family of conics yz-kx2 = 0, defined through the sides of the triangle ABC (x = 0 corresponding to BC, y=0 to CA and z = 0 to AB).

[1] From the general theory, we know, since AB, AC are tangent to the conic, that AM must be a diameter of the conic conjugate to BC, M being the middle of BC. Hence the center of the conic lies on the median line AM.
[2] The directions of the asymptotes are immediately constructible from the data. This because they define a right-angled triangle symmetric about the middle M of BC.
[3] To find these directions take an arbitrary point D on BC and points E, G such that ME = MG = MD as shown. The so constructed triangle DEG is right-angled at G and its sides are parallel to the asymptotes of the hyperbola. All right-angled triangles constructible in this way are pairwise similar.
[4] Extend the orthogonal sides of the constructed triangle to cut one of the two sides of ABC tangent to the hyperbola under construction. Thus, create triangle HGI. All triangles constructed in this way are pairwise similar, hence the direction of their median GK from G is also determined by one of them.
[5] Knowing the center O and the asymptotics, consider their intersections L, N with side AC. Triangle OLN is an asymptotic one of the hyperbola.
[6] Since all asymptotic triangles have the same area and the axes are known (bisect the asymptotes in this case), construct the asymptotic triangle OPQ as a right-angled isosceles with given area.
The vertex of the hyperbola is on the middle X of PQ, and this completes the determination of the hyperbola, in terms of the triangle's data.
See BitangentRectangularMember2.html for a further discussion on this topic leading to a simpler method to define the hyperbola.