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' = 0

4y^3 y' - 2ayy' + 4x^2 yy' - bxy' + 4x^3 + 2ax + 4xy^2 - by = 0

y' = -(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 y

with 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 on

4x^3 + 2ax + 4xy^2 - by = 0 to get y in terms of x and then

substitue y = f(x) into the first equation and solve for x.

That's a heafty task. Next, get values for y and check to see

if they satisfy the addional requirement and then if they're

maximal, 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

>