Expert: Paul Klarreich - 3/11/2007

Question
Hi there, how could i formulate this mathematically?
There is an equilateral hyperbole defined by xy=1. 3 points A, B, C belong to this hyperble. How to show that the othocenter of the triangle ABC also belongs to the hyperbole?

Answer
Questioner:   Sean
Category:  Advanced Math
 
Subject:  Algebraic curves
Question:  Hi there, how could i formulate this mathematically?
There is an equilateral hyperbole

>> No, no.  Hyperbole is what we get from the Bush administration.  This is a hyperbola, with an A.


defined by xy=1. 3 points A, B, C belong to this hyperbola. How to show that the othocenter of the triangle ABC also belongs to the hyperbola?
...........................................
Hi, Sean,

The orthocenter is the point of intersection of the three altitudes of the triangle.  If A,B,C are the vertices of your triangle, you find the (equation of) the altitude by this process:

Altitude to AB:  This is a line perpendicular to AB passing through C.

1. Get the slope of AB = m. (Easy - use slope formula.)
2. Take its negative reciprocal, -1/m
3. Use that slope and the point C to get the equation.

Now the altitude to BC is a line perpendicular to BC from A.  Interestingly, the 'orthocenter' of a triangle is not always INSIDE the triangle.

Let's try it.  We need some good notation for the points.  Remember, they are on the hyperbola.

Let the points be  A(a,1/a), B(b,1/b), C(c,1/c)
.....................................
Altitude to AB, from C.  USUAL WARNING ABOUT COURIER FONT.
   1/b - 1/a     a - b       -1
m = --------- = ---------- = ----
    b - a      ab(b - a)     ab

Neg recip:  -1/m = ab

Use C(c,1/c) now:

y - 1/c = ab(x - c)
cy - 1 = abc(x - c)
cy - 1 = abcx - abc^2
....................................
Altitude to BC, from A.
   1/c - 1/b     b - c      -1
m = --------- = --------- = ----
     c - b     bc(c - b)    bc

Neg recip: -1/m = bc

Use A(a,1/a) now:

y - 1/a = bc(x - a)
ay - 1 = abc(x - a)
ay - 1 = abcx - a^2bc
..................................
Now we have two equations to solve:

ay - 1 = abcx - a^2bc  (I)
cy - 1 = abcx - abc^2  (II)

Solve by eliminating y.

acy - c = abc^2 x - a^2bc^2    (c times equation I)
acy - a = a^2bc x - a^2bc^2    (a times equation II)

Subtract:  (aII - cI)

(c - a) = (a^2bc - abc^2)x
     (c - a)
x = ---------------
   (a^2bc - abc^2)

     (c - a)
x = ------------
   abc(a - c)

   -1
x = ----
   abc

Substitute back into I:

ay - 1 = abcx - a^2bc  (I), with  x = -1/abc

ay - 1 = -1 - a^2bc

ay  = - a^2bc

y = - abc

OK, then, here are your coordinates:

x = -1/abc,  and  y = - abc.

AND, Ladies and Gentlemen:

xy = (-1/abc)(-abc) = 1,

so (x,y) is on the hyperbola.