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Asymptotes of circumhyperbola

To find the asymptotes of a circumhyperbola, pespector P = (l:m:n), we
1. Find the intersection of the dual of P with the Steiner ellipse.
2. Find the infinite points on the hyperbola as the isotomic conjugates of the points in step 1
3. Find the equation of the lines from the center of the hyperbola to the infinite points.

The computations are hairy so they are done here using Mathematica.

Intersection of generic line and Steiner ellipse

This finds the intersection of the dual of P = (l:m:n) with the Steiner ellipse. The dual of P is the line lx + my + nz = 0.
Click on link to see the Mathematica compuation.

Solve[{ {l, m, n} . {x, y, z} == 0, y z + z x + x y == 0, x + y + z == 1}, {x, y, z}]//FullSimplify

write as points

clear denominator, same in all terms.

under square root; we call this

The point of intersection is this formula.
Here Z^2 = -2 l m+-2 l n-2 m n+

Find the asymptotes

The infinite points on the asymptotes are the isotomic conjugates of the intersections with the Steiner ellipse. The line between the infinite point and the center of the conic is an asymptote.

$asy1 = intersection[iso[{m^2 + n^2 - l (m + n) + m Z - n Z, l^2 - l (m + Z) + n (-m + n + Z), l^2 + m^2 - l n - m n + (l - m) Z}], {l (m + n - l), m (n + l - m), n (l + m - n)}]$

Since this factor = (m-n), we can simplify the asymptote equation to.

$asy1 = {l (l m - m^2 + l n - n^2 - m Z + n Z)^2, m (l^2 - l m - m n + n^2 - l Z + n Z)^2, n (l^2 + m^2 - l n - m n + l Z - m Z)^2 }$

${l (l m - m^2 + l n - n^2 - m Z + n Z)^2, m (l^2 - l m - m n + n^2 - l Z + n Z)^2, n (l^2 + m^2 - l n - m n + l Z - m Z)^2}$

We find the other asymptote

${l (l m - m^2 + l n - n^2 + m Z - n Z)^2, m (l^2 - l m - m n + n^2 + l Z - n Z)^2, n (l^2 + m^2 - l n - m n - l Z + m Z)^2}$

Hence the equation of an asymptote is

${l (m^2 - l n + n^2 - l m + (m - n) Z)^2, m (n^2 - l m + l^2 - m n + (n - l) Z)^2, n (l^2 - m n + m^2 - l n + (l - m) Z)^2}$

${ l (m^2 - l n + n^2 - l m + (m - n) Z)^2, m (n^2 - l m + l^2 - m n + (n - l) Z)^2, n (l^2 - m n + m^2 - l n + (l - m) Z)^2 }$

(1)

Now we show that the asymptotes meet at the center; this is a check.

${-2 l (l - m - n) (l^2 - l m + m^2 - l n - m n + n^2), 2 m (l - m + n) (l^2 - l m + m^2 - l n - m n + n^2), 2 (l + m - n) n (l^2 - l m + m^2 - l n - m n + n^2)}$

Check!

The asymptotes are isotomic

asy[[1]] is the first coordinate of asy, the asymptote.

$((asy1[[1]] asy2[[1]]//Expand)/.{Z^2->l^2 - 2 l m + m^2 - 2 l n - 2 m n + n^2, Z^4→ (l^2 - 2 l m + m^2 - 2 l n - 2 m n + n^2)^2})//Factor$

Check!

Perspector at infinity

If P is at infinity, it has the form (m-n:n-l:l-m}. Evaluate

$l^2 - 2 l m + m^2 - 2 l n - 2 m n + n^2/.{l→m - n, m→n - l, n→ l - m}//Expand//Factor$

Hence Z becomes 2Q.
Q =

$(l^2 - l m - m n + n^2/.{l→m - n, m→n - l, n→ l - m})//Expand//Factor$

Hence = 2 so that the factor +(n-l)Z becomes 2+ (n+l-2m)2Q = Q + (n+l-2m).
Hence the asymptote is :(n-l)( n+l-2m :

Let's see if they are isotomic by multiplying the second coordinates of both asymptotes

$(((m - n) (Q - 2 l + m + n )^2 (m - n) (Q + 2 l - m - n )^2//Expand)/.{Q^2->l^2 - l m + m^2 - l n - m n + n^2, Q^4→ (l^2 - l m + m^2 - l n - m n + n^2)^2})//Expand//Factor$

They are!

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Created by Mathematica (September 5, 2006)