Asked by Ivy
Write the equation of the tangent at (2,2) to the curve x^2-2xy+y^2+2x+y-6=0
Can anyone please give me some ideas to do it?Thanks!
Can anyone please give me some ideas to do it?Thanks!
Answers
Answered by
Reiny
first you have to differentiate implicitly with respect to x
2x - 2xdy/dx - 2y + 2ydy/dx + 2 + dy/dx = 0
now solve this for dy/dx by using common factors
dy/dx(2y - 2x + 1) = -2x + 2y + 2
dy/dx = (-2x+2y+2)/(2y-2x+1)
now sub in x = 2 from your point (2,2) to get the slope of the tangent.
from there it should be easy, let me know what you got.
2x - 2xdy/dx - 2y + 2ydy/dx + 2 + dy/dx = 0
now solve this for dy/dx by using common factors
dy/dx(2y - 2x + 1) = -2x + 2y + 2
dy/dx = (-2x+2y+2)/(2y-2x+1)
now sub in x = 2 from your point (2,2) to get the slope of the tangent.
from there it should be easy, let me know what you got.
Answered by
Ivy
Is it y=2x-2?
Answered by
Reiny
correct
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