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Math Help - Find an equation of the line tangent

  1. #1
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    Find an equation of the line tangent

    Find an equation of the line tangent to the curve 2y^{2}-x^{2}=1 at the point (7,-5)
    I tried to solve it but can't get the right answer, can you help me check it out?

    2y^{2}-x^{2}=1

    Derivative

    4yy' - 2x = 0
    y'=\frac{x}{2y}
    y'(7)=\frac{-7}{10}
    y=mx+b => -5=\frac{-7}{10}7+b=>b=\frac{-1}{10}

    The equation is

    y=\frac{-7}{10}x-\frac{1}{10}

    But the right answer is

    y=\frac{-7}{10}x-\frac{17}{2}

    Where is my error?
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  2. #2
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    Re: Find an equation of the line tangent

    If I'm not mistaken your answer is correct.
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  3. #3
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    Re: Find an equation of the line tangent

    I agree with Cesc1 - your answer is correct.

    - Hollywood
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Find an equation of the line tangent

    Quote Originally Posted by shiho View Post
    Find an equation of the line tangent to the curve 2y^{2}-x^{2}=1 at the point (7,-5)
    I tried to solve it but can't get the right answer, can you help me check it out?

    2y^{2}-x^{2}=1

    Derivative

    4yy' - 2x = 0
    y'=\frac{x}{2y}
    y'(7)=\frac{-7}{10}
    y=mx+b => -5=\frac{-7}{10}7+b=>b=\frac{-1}{10}

    The equation is

    y=\frac{-7}{10}x-\frac{1}{10}

    But the right answer is

    y=\frac{-7}{10}x-\frac{17}{2}

    Where is my error?
    Let's see... suppose the "right answer" is right...
    Then the point (7, -5) should be on this tangent line.

    So:
    y=\frac{-7}{10}x-\frac{17}{2}

    -5=\frac{-7}{10}\cdot 7-\frac{17}{2}

    -5=\frac{-49}{10}-\frac{17}{2} = -13.4

    Contradiction!

    So the "right answer" is wrong, or at least it does not belong to the given problem.
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  5. #5
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    Re: Find an equation of the line tangent

    ah thanks for your explanation
    maybe this time the answer paper is wrong xD
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