
Re: Building an Equation to find (Maximum Y) ie Highest Point on a curve!
Posted:
Sep 12, 2013 9:15 PM


>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 = 0 as the equation to be merged with the xyaltered 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.
Thank you for replying Leon
Mervyn

