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

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Answer 1 · 2016-11-11T08:37:17

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)#

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):}#