Mathematics

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

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  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

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