```Date: Sep 13, 2013 4:58 AM
Author: Thomas Nordhaus
Subject: Re: Building an Equation to find (Maximum Y) ie Highest Point on<br> a curve!

Am 13.09.2013 03:15, schrieb mervynmccrabbe@gmail.com:>> Just interchange x and y, then get Max y for the new equation.>> x^4  +  y^4  +  A(x^2)  -   A(y^2)   +  2(x^2)(y^2)  -  Bxy  +  C   =  0>> By substituting x for y the above equation becomes :-> x^4  +  y^4  -  A(x^2)  +   A(y^2)   +  2(x^2)(y^2)  -  Bxy  +  C   =  0>> So presumably the new dy/dx becomes:-> dy/dx = -(4x^3 - 2ax + 4xy^2 - by)/(4y^3  + 2ay + 4x^2 y - bx)>> and in turn giving> 4x^3 - 2ax  + 4x(y^2) - by = 04x^3 - 2Ax + 4x(y^2) - By = 0> as the equation to be merged with the xy-altered equation:-> x^4  +  y^4  -  A(x^2)  +   A(y^2)   +  2(x^2)(y^2)  -  Bxy  +  C   =  0>> Even if i'm right so far, I am again lost in finding the equivalent of> quasi's solution to the original equation.Look closer: The only difference in the set of equations is that A is replaced by -A. Now look at the coefficients d8, d6, d4, d2, d0 from quasi's equation for y. What happens when you replace A by -A:d8 -> d8 (nothing changes because of only even powers in A)d6 -> -d6 (switch sign because of only odd powers in A)d4 -> d4d2 -> -d2d0 -> d0.So your equation for x becomes:d8*x^8 - d6*x^6 + d4*x^4 - d2*x^2 + d0 = 0.This is the power of symmetry ;-)-- Thomas Nordhaus
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