## Hyperbola Generation through equal segments

Let a triangle ABC and point B' on side BC. Take a variable point A' on AC and consider the middles C' of BB' and E of AA'. The variable lines {C'E,B'A'} intersect at a point K describing a hyperbola passing through points {A,B',C'} and having its asymptotes parallel to sides {AB, AC} of the triangle, consequently its axes being parallel to the bisectors of angle A.

Take the projective basis {A,B',C',G}, G being the centroid of triangle AB'C'.
C = B'-kC' (fixed) and A' = A + rC = A + r(B'-kC') = A+rB'-(kr)C' (variable).
To find the middle E of AA' identify it with the harmonic conjugate of the point at infinity of AA'.
AA' : A + tA' = A + t(A+rB'-(kr)C') = (1+t)A + (tr)B' -(tkr)C', intersect with (x+y+z=0).
(1+t)+(tr)-(tkr) = 0 => t = 1/(kr-r-1) (*). Thus the middle E of AA' is given by:
E = A - [1/(kr-r-1)]A' = (1-t)A - (tr)B' + (tkr)C'.
Line B'A' has coefficients : (0,1,0) x (1, r, -kr) = (kr, 0, 1).
Line C'E has coefficients : (0,0,1) x ( 1-t, -tr, tkr ) = (tr, 1-t, 0).
The intersection point K has coordinates (kr, 0, 1) x (tr, 1-t, 0) = ( t-1, tr, kr(1-t) ).
Setting the last triple equal to (mx, my, mz) and eliminating {m,r,t} using also (*) we find the equation:
(k-1)yz + zx + (2k)xy = 0.
This is the equation of a conic passing through the vertices of triangle AB'C'.

To find its intersection with the line at infinity x+y+z=0, replace in the equation z = -(x+y) to get
(1-k)y2 + kxy - x2 = 0 => (1-k)w2 + kw -1 = 0 (where w = y/x).
The discriminant being d = k2 + 4(1-k) = (k-2)2, we get the solutions
w1 = 1/(k-1) and w2 = 1.
The value k=2 giving two equal solutions would correspond to point C such that CB'/CC' = k = 2 (see BarycentricCoordinates2.html ) identifying it with B which is impossible according to the definition of the configuration. Thus there are always two points on the line at infinity and the conic is a hyperbola.

The asymptotes are the tangents at the points at infinity (1, w, -1-w), which for w=1 gives (1,1,-2) and for w=1/(k-1) gives (k-1,1,-k). On the other side B=B'-2C' determines line AB with coefficients (1,0,0)x(0,1,-2)=(0,2,1) and corresponding point at infinity of the line (0,2,1)x(1,1,1)=(1,1,-2).
Analogously C=B'-kC' determines line AC with coefficients (1,0,0)x(0,1,-k)=(0,k,1) and corresponding point at infinity (0,k,1)x(1,1,1)=(k-1,1,-k).
Thus it is seen that lines {AB,AC} determine the same points at infinity with the points at infinity of the hyperbola hence are parallel to the asymptotes.

Remark-1 The fact that the above locus is a conic follows easily by reducing the case to a Chasles-Steiner generation of conics (see Chasles_Steiner.html ). For this consider the two bundles of lines at {B',C'}. Taking x=AE on line AC and 2x=AA' we see that the intersection point K results by a homographic relation between these two bundles. Thus K describes a conic passing through {B',C'} and the special point A for this relation, resulting for x=0. The calculations are only needed in order to obtain a more detailed picture of the conic.
Another similar case is handled in ChaslesSteinerExample.html .
Remark-2 Having calculated the points at infinity of the hyperbola we can find also the tangents there i.e. the asymptotes and their intersection points with the sides of ABC, thus giving a complete picture of the location of the conic.

### See Also

BarycentricCoordinates2.html
Chasles_Steiner.html
ChaslesSteinerExample.html
Hyperbola.html

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