It is natural to consider the coordinate system with axes the lines {OB,OC}.

Then calculate first the coefficients of line DX and its intersections with these axes.

The coefficients of the line EF are given by the vector product (x,y,1) x (a,b,1) = (y-b, a-x, bx-ay).

Its intersections with the axes are

E(0,y'), with y' = (ay-bx)/(a-x), and

F(x',0), with x' = (ay-bx)/(y-b).

The basic relation results from the similar triangles:

y/(y-y') = r <=> y = r(y-y') <=> y = r(y - (ay-bx)/(a-x)) <=> y = r(bx-xy)/(a-x) <=>

xy = Ux + Vy, with U = rb/(r-1) and V = a/(1-r).

Translate to the coordinate system (x*,y*) such that

x = x* + d

xy = x*y* + x*d

Selecting d

x*y* = UV.

This short calculation proves the claim and delivers also the basis of a precise drawing of the hyperbola. The results of HyperbolaProperty.html (*) can be easily deduced from these here. In fact, it is easy now to show that the locus of the symmetric of O with respect to X is also a hyperbola. The problem in (*) corresponds then to the special case r = -1, making X the middle of EF.

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