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Math Help - Find the equation,

  1. #1
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    Find the equation,

    I thought this was a lil to simple and think my answer might be wrong :/

    Determine the equation of the tangent to the curve at the given point

    (x+2)^2+(y-3)^2=29 P(3,5)
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  2. #2
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    Quote Originally Posted by tnkfub
    I thought this was a lil to simple and think my answer might be wrong :/

    Determine the equation of the tangent to the curve at the given point

    (x+2)^2+(y-3)^2=29 P(3,5)
    The slope of the tangent line is dy/dx.

    (x+2)^2 +(y-3)^2 = 29
    Differentiate both sides with respect to x,
    2(x+2) +2(y-3)(dy/dx) = 0
    dy/dx = -(x+2)/(y-3) -----------***

    At point (3,5),
    dy/dx = -(3+2)/(5-3) = -5/2
    Using the point-slope form of the equation of the tangent line,
    y-5 = -(5/2)(x-3)

    y = -(5/2)(x-3) +5
    y = -(5/2)x +15/2 +5
    y = -(5/2)x +25/2 ----------answer.
    Or,
    2y = -5x +25
    5x +2y -25 = 0 -------answer.
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