Calculus Question. Expert help.?

Question

Consider the curve defined by 2y^3+6x^2y-12x^2+6y=1

I found dy/dx= 4x-2xy/x^2+y^2+1

Now it wants me to write an equation of each horizontal tangent line to the curve.

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

Answer 1

taking the derivative of both sides

6y^2*y´+12x*y + 6x^2y´-24x +6*y´=0 (a)

If the tangent is horizontal y´=0 so

12x*y -24 x=0 12x*(y-2) =0 so x= 0 and 2y^3+6y-1=0 which has onli one root between 0 and 1(if rational p/q,p=+1 q=+ 2which doesn´t verify)

y=2 and there is no x

y=-x is tangent and putting y´=-1 in (a) and x =-y you get

-6y^2+6(-2y^2-y^2) +24y-6=0 ==>-24y^2 +24y -6 = 0

4y^2 -4y+1=0 so y=1/2 x=-1/2