Question

use implicit differentiation to find the point where the parabola defined by x^2-2xy+y^2+4x-8y+20=0

Answer 1

Solution for x^2+y^2-4x+8y+20=0 equation:

Simplifying
x2 + y2 + -4x + 8y + 20 = 0

Reorder the terms:
20 + -4x + x2 + 8y + y2 = 0

Solving
20 + -4x + x2 + 8y + y2 = 0

Solving for variable 'x'.

The solution to this equation could not be determined.

Answer 2

Your original question is incomplete.
However, for x^2-2xy+y^2+4x-8y+20=0
2x - 2x(dy/dx) - 2y + 2y(dy/dx) + 4 - 8dy/dx = 0
dy/dx(-2x + 2y - 8) = -2x + 2y - 4
dy/dx = (-2x + 2y - 8)/(-2x + 2y - 4)
= (x - y + 4)/(x - y + 2)

btw, this is what your graph looks like
https://www.wolframalpha.com/input/?i=plot+x%5E2-2xy%2By%5E2%2B4x-8y%2B20%3D0

Now what is your question?

Answer 3

I already showed you how to find where the parabola has vertical and horizontal tangents. What did you not like about my solution?

Answer 4

oops - that may have been someone else. Try

https://www.jiskha.com/questions/1805378/Use-implicit-differentiation-to-find-the-points-where-the-parabola-defined-by-x-2-2xy-y-2-8x-4y-20