# Finding the positive x-value on a hyperbola

1. Mar 11, 2009

### Emethyst

1. The problem statement, all variables and given/known data
The curve y^2-3xy+2x^2=4 is a hyperbola with axes rotated from the standard position. Use Newton's Method to find the positive x-value to four decimal places for the point on the hyperbola where y=1.

2. Relevant equations
Newton's Method

3. The attempt at a solution
I found the first part of Newton's Method by finding the derivative of the equation given, but I don't know how to find f(x) to finish of the formula. I've figured that you can simply plug the y-value into the given equation, make it equal to zero, and then plug it in for f(x), but then I do not know the starting value to use for x. I know how to use Newton's Method and find the derivative, but for this question I just don't know how to find f(x) and the starting x-value needed to solve for the answer. Any help you guys can give would be greatly appreciated, thanks in advance.

2. Mar 11, 2009

### lanedance

hi emethyst

first substitute y = 1 into your equation and rearrange for

so it looks like
f(x) = 0
and you want to find x that satisfies the equation

this will be a quadratic so you could in fact solve it, and use the quadratic equation as a check

then think about a negative parabola (which is what f(x) is...) where would you want to pick a point so that you newton iterations find the positive x value & don't over shoot in the process

doing an approximate curve sketch might help...
what is the turning point, and where does the curve intersect f(x) axis when x is zero, should be enough to pick a reasonable point

3. Mar 12, 2009

### Emethyst

Thanks for all the help lanedance, I can say I successfully solved that question now