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

1 Answer
Sep 10, 2016

y = pm 6
x = pm6

Explanation:

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