Consider three fixed points {A,B,C} and a fixed line AD trhough A. On line AD choose points {B',C'} such that B'A/B'C' = k is a fixed constant. The intersection points P of the variable lines {BB',CC'} generate a hyperbola.

That this is a conic follows immediately by taking the origin of coordinates on AD to be at A and setting AB'=x, AC'=y, this means that -x/(y-x)=k => y=((k-1)/k)x, which is a very simple

To see that this is a hyperbola suffices to notice that it intersects the line at infinity at two distinct points. The first is determined by the direction of the line AD, since B' going to infinity implies that C' does the same and lines {BB',CC'} become parallel to AD.

The other point at infinity lying on the conic is found as follows. Note first that since B'A/B'C' is constant the parallel to BB' from C' intersects AB at a fixed point A

The proof of the property can be based on section-1. After it, forgetting for the moment the given hyperbola c, we determine another hyperbola c' describing the locus of a point P as in section-1. For this take an arbitrary point P

In these coordinates point D (fixed) on BC has the expression D=B+pC. The variable points B'=D+mA=mA+B+pC, C'=D+nA=nA+B+pC are related by having a constant ratio B'A/B'C'=k.

By the general theory the locus of points P is a conic passing through {A,B,C} hence having the form :

Myz + Nzx + Pxy =0 (*).

In addition the curve passes through two points at infinity determined by the given conditions. The first is the point at infinity determined by line AD. This point is obtained as B' tends to this point at infinity along AD.

The second point is the point at infinity of line A

[1] Line AD : has coefficients given by the vector product (1,0,0)x(0,1,p)=(0,-p,1), with point at infinity (0,-p,1)x(1,1,1) =

(-1-p, 1, p).

[2] The second point at infinity needs point A

Then line CA

Introducing these into (*) we obtain the equations for {M,N,P}:

(-p)M +p(1+p)N + (1+p)P = 0,

k(k-1)M +kN + (1-k)P = 0.

Obviously (M,N,P) is a multiple of the vector product (-p,p(1+p), (1+p)) x (k(k-1), k, 1-k), which leads easily to the conic equation:

(1+p)yz + (1-k)zx + (pk)xy = 0.

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