Let A, B, and C be three points on the curve xy = 1 (which is a hyperbola). Prove that the orthocenter of triangle ABC also lies on the curve xy = 1.
Let the points be A =(1/a,a), B = (1/b, b) and C = (1/c, c)
The slope between B and C = [c - b] / [1/c - 1/b] = [c - b] / [b - c] / bc = -bc
So..... the equation of a perpendicular line to BC passing through A will be
y = (1/bc) (x - 1/a) + a (1)
Likewise......the equation through B perpendicular to AC is
y = (1/ac)(x - 1/b) + b (2)
And the equation through C perpendicular to AB is
y = (1/ab)(x - 1/c) + c (3)
Setting (1) = (2) we have
(1/bc) (x - 1/a) + a = (1/ac)(x - 1/b) + b simplify
x/bc - 1/abc + a = x/ac - 1/abc + b
x/bc + a = x/ac + b
x ( 1/bc - 1/ac) = b - a
x [ ( a - b] / (abc) ] = b - a
x [ (a - b) / (abc) ] = - (a - b)
x = -abc
Sub this into (3)
y = (1/ab) (-abc - 1/c) + c
y = -abc/ ab - 1/abc + c
y = - c - 1/abc + c
y = -1/abc
So.....the orthocenter is at ( x, 1/x) = ( -abc, - 1/abc)
To prove this point is on the function xy = 1, we have
( x ) ( y ) =
(-abc) (-1/abc) =
(-1) (-1) = 1
As an aside, the triangle formed is obtuse, so the orthocenter will fall outside the triangle
Here is an example.....points A,B, C are on xy = 1 .....D is the orthocenter