+0  
 
0
288
1
avatar

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.

Guest Apr 17, 2017
Sort: 

1+0 Answers

 #1
avatar+77125 
+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

 

 

cool cool cool

CPhill  Apr 18, 2017
edited by CPhill  Apr 18, 2017
edited by CPhill  Apr 18, 2017
edited by CPhill  Apr 18, 2017

17 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details