Find the equation of a line tangent to te curve xy = sqrt(xy - x) + 1 at the point (1, 2).

Answer 1

xy = (xy - x)^(1/2)
x dy/dx + y = (1/2)(xy - x)^(-1/2) (xdy/dx + y - 1)
so at (1,2)
dy/dx + 2 = (1/2)(2 - 1)^(-1/2) (dy/dx + 2 - 1)
times 2

2dy/dx + 4 = (1)(dy/dx + 1)
dy/dx = 1-4 = -3

so tangent is y = -3x + b
with (1,2) lying on it
2 = -3+b
b=5

tangent equation:
y = -3x + 5

check my arithmetic