Find the point(s) (if any) of horizontal tangent lines: x^2+xy+y^2=6

Answer 1

2x + y + xy' + 2yy' = 0
y'(x+2y) = -(2x+y)

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

So, y'=0 when 2x+y = 0
So, if y = -2x,

x^2-2x^2+4x^2 = 6
3x^2 = 6
x = √2
So, y = -2√2

At (√2,-2√2) and (-√2,2√2) the tangent is horizontal.

Check:

http://www.wolframalpha.com/input/?i=plot+x^2%2Bxy%2By^2%3D6+and+y%3D2%E2%88%9A2+and+y%3D-2%E2%88%9A2