```Date: Sep 12, 2013 2:41 AM
Author: William Elliot
Subject: Re: Equation to find Highest Point on a curve!

On Wed, 11 Sep 2013, mervynmccrabbe@gmail.com wrote:> x^4  +  y^4  +  A(x^2)  -   A(y^2)   +  2(x^2)(y^2)  -  Bxy  +  C   =  0> > If I have correctly evaluated > dy/dx to be   =      [4(x^3) + 2Ax  + 4x(y^2)  -  By]You haven't.  Letting y' = dy/dx,4x^3 + 4y^3 y' + 2ax - 2ayy' + 4xy^2 + 4x^2 yy' - by - bxy' = 04y^3 y' - 2ayy' + 4x^2 yy' - bxy' + 4x^3 + 2ax + 4xy^2 - by  = 0y' = -(4x^3 + 2ax + 4xy^2 - by)/(4y^3  - 2ay + 4x^2 y - bx)> and if it is proper to set this then equal to zero> to give a new equation that could be merged with the original to get rid > of the cumbersome XY terms - then that i failed to do.What?  To find extreme values of y, set y' = 0.That gives you two equations to solve for x and ywith the additional requirement that 	4y^3  - 2ay + 4x^2 y - bx /= 0>  I've tried completing squares etc but can not get rid of composite XY > terms.> Complete the square on4x^3 + 2ax + 4xy^2 - by = 0 to get y in terms of x and thensubstitue y = f(x) into the first equation and solve for x.That's a heafty task.  Next, get values for y and check to seeif they satisfy the addional requirement and then if they'remaximal, miminal or points of inflection.>  If I could eliminate X and get a generalised Y-Only equation >  then I could manage the rest.> >  Any help would be appreciated> >  Mervyn Mc Crabbe>
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