Find the point(s) (if any) of horizontal tangent lines for the equation x^2+xy+y^2=6. If none exist, why?

1 Answer
Nov 11, 2016

graph{(x^2+xy+y^2-6)((x-sqrt2)^2+(y+2sqrt2)^2-0.01)((x+sqrt2)^2+(y-2sqrt2)^2-0.01)=0 [-8.89, 8.89, -4.444, 4.445]}

P_1=(sqrt2, -2sqrt2)
P_2=(-sqrt2, +2sqrt2)

Explanation:

Implicitly differentiating over x the equation we get (using y'=dy/dx)

2x+y+xy'+2yy'=0

So

2x+y=-(x+2y)y'

y'=-(2x+y)/(x+2y)

So y'=0 only when y=-2x (excluding y=0 when y'=-2)

Now intersect with the equation

{(y=-2x),(x^2+xy+y^2=6):}
{(y=-2x),(x^2-2x^2+4x^2=6):}
{(y=-2x),(3x^2=6):}
{(y=-2x),(x=+-sqrt2):}