Posted by **Stewart** on Tuesday, June 5, 2012 at 6:58pm.

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

- Equation of the Tangent Line -
**Reiny**, Tuesday, June 5, 2012 at 8:02pm
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

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