Calculus...URGENT..show steps please

Use implicit differentiation to find the points where the parabola defined by
x^{2}-2xy+y^{2}-6x+2y+17 = 0
has horizontal and vertical tangent lines. List your answers as points in the form (a,b).

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  1. 2xdx-2y dx-2xdy + 2y dy -6dx+2dy=0

    for horizontal lines, dy/dx=0
    2x-2y-6)/(-2x+2y+2)=0
    or y=x-3

    for vertical tangent, dy/dx=undifined
    or -2x+2y+2=0
    y=x-2

    now solve the points on the parabola. for horizontal lines, substutute x-2 for y in the given equation, solve.

    Then, do the same for vertical lines.

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    bobpursley
  2. The "solving" becomes a bit easier if you change the original equation to

    (x-y)^2 - 6x + 2y + 17 = 0

    sub in : y = x - 3
    (x - (x-3))^2 - 6x + 2(x-3) + 17 = 0
    9 - 6x + 2x - 6 + 17 = 0
    -4x = -20
    x = 5
    then y = 2
    the horizontal asymptote touches at (5,2)

    I will do the vertical asymptote without the simplification:

    x^2 - 2x(x-2) + (x-2)^2 - 6x + 2(x-2) + 17 = 0
    x^2 - 2x^2 + 4x + x^2 - 4x + 4 - 6x + 2x-4 + 17 = 0
    -4x = -17
    x = 17/4
    then y = 17/4 - 2 = 9/4
    the vertical asymptote touches at (17/4 , 9/4)

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