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Posted by 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 - , 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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