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Thread: parabola

  1. #1
    Super Member dhiab's Avatar
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    parabola

    Find the point P on the parabola y = x closest to the point (3, 0).

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  2. #2
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    Re: parabola

    $L=d^2 = (x-3)^2 + (x^2-0)^2 = (x-3)^2 + x^4$

    $\dfrac{dL}{dx} = 2(x-3) + 4x^3$

    $\dfrac{dL}{dx} = 0$

    $2(x-3)+4x^3=0$

    $2(3-x) = 4x^3$

    $3-x=2x^3$

    This has 1 real solution

    $x=1$

    $\dfrac{d^2L}{dL^2} = \left . 2+12x^2\right |_{x=1} = 14>0$

    So this is a minimum.

    The point is $(1,1)$
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  3. #3
    Super Member dhiab's Avatar
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    Re: parabola

    hello thank you the point p is: p(1 ; 5) non (1; 1)
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    Re: parabola

    Quote Originally Posted by dhiab View Post
    hello thank you the point p is: p(1 ; 5) non (1; 1)
    That is absolutely impossible. The point $(1,5)$ is not even on the parabola.
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  5. #5
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    Re: parabola

    Quote Originally Posted by Plato View Post
    The point (1,5) is not even..
    Agree: 1 and 5 are odd numbers
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