Let A and B be two points on the hyperbola xy=1, and let C be the reflection of B through the origin.
Let Gamma be the circumcircle of triangle ABC and let A' be the point on Gamma diametrically opposite A. Show that A' is also on the hyperbola xy=1.
2 answers
Nvrm I solved it.
the hyperbola xy=1 has two branches, symmetric about the origin, since (-x)(-y) = 1
So, if A = (x,y), then A'=(-x,-y) and is also on the hyperbola.
There is no need to introduce any circle, since A' is directly opposite A.
So, if A = (x,y), then A'=(-x,-y) and is also on the hyperbola.
There is no need to introduce any circle, since A' is directly opposite A.