How do you find horizontal and vertical tangent lines after using implicit differentiation of #x^2+xy+y^2=27#?

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Answer 1 · 2016-09-10T21:24:36

#y = pm 6#
#x = pm6#

Given #f(x,y)=x^2+xy+y^2-27=0#

#df=f_x dx + f_y dy = 0#

so

#dy/dx = - f_x/(f_y) = (2x+y)/(2y+x)#

The horizontal tangent lines have #f_x = 0->x = -y/2# and the vertical tangent lines have #f_y = 0->x = -2y#

So for horizontals

#f(-y/2,y) = y^2/4-2y^2+y^2-27=0->y=pm6#

and for verticals

#f(x,-x/2) = x^2-x^2/2+x^2/4 - 27=0->x=pm 6#