Use implicit differentiation to find the points where the parabola defined by

Question

x2−2xy+y2−4y+4=0
has horizontal and vertical tangent lines.

The parabola has horizontal tangent lines at the point(s)
(x-y)/x-y+2
.

The parabola has vertical tangent lines at the point(s)

Answer 1

Well, you got that y' = (x-y)/(x-y+2)
If x=y, then y' = 0/2 = 0 so horizontal tangent
If x-y = -2 then y' = -2/0 so vertical tangent

So, where on the parabola is x=y? When
x^2−2x^2+x^2−4x+4=0
x = 1
So, at (1,1) the tangent is horizontal.
similarly, at (0,2) the tangent is vertical