Question

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).

Answer 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.

Answer 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)