SOLUTION: Find the equation of the tangent line to the curve (x-y)^2=2x+1 at the point (4,1).

Question 1010264: Find the equation of the tangent line to the curve (x-y)^2=2x+1 at the point (4,1).
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the equation of the tangent line to the curve (x-y)^2=2x+1 at the point (4,1).
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(x-y)^2=2x+1
x^2 - 2xy + y^2 = 2x+1
Differentiate implicitly
2xdx - 2ydx - 2xdy + 2ydy = 2dx
xdx - ydx - xdy + ydy - dx = 0
(x-y-1)dx - (x - y)dy = 0
(x-y-1)dx = (x - y)dy
dy/dx = (x-y-1)/(x-y)
@ (4,1) dy/dx = 2/3
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y-1 = (2/3)(x-4)

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