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The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y. What is the equation of the curve if (3, 1) is a point on the curve?

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  1. slope = x/y = 3/1 = 3

    so the equation of the tangent at the point (3,1) is
    y-1 = 3(x-3)
    y = 3x -8 or 3x - y - 8 = 0

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  2. y' = x/y
    you can recognize that as an hyperbola

    x^2-y^2 = a^2
    since (3,1) is on the curve,
    9-1 = a^2

    x^2-y^2 = 8

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  3. Maybe I should become a politician.
    Never actually answer the question, rather give an answer to some other question.
    Thanks for the catch Steve, that is two for one evening, I should get more sleep.

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