Implicitly difined curve, fine point with given slope.

[saldana787]

Homework Statement

Here is the full question.
Consider a curve defined by 2y^3 + 6x^2y - 12x^2 + 6y = 1 and dy/dx = (4x-2xy)/(x^2+y^2+1).

The line through the origin with slope -1 is tangent to the curve at point P. Find the x - coordinate and y - coordinate of point P.

Homework Equations

2y^3 + 6x^2y - 12x^2 + 6y = 1
dy/dx = (4x-2xy)/(x^2+y^2+1)

The Attempt at a Solution

The equation for the line through (o,o) with slope -1 which I found to be y = -x. I tried to plug -x back into original equation but came up with a weird expression for y.

Thanks for any help in advance.

 

[Dick]

Weird in what way? Sure, the tangent line is y=(-x). You also have dy/dx=(-1), right? What's so weird about that? You'll have to tell us.