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Math Help - Normal lines on a parabola

  1. #1
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    Normal lines

    Thanks!
    Last edited by Blasnmt23; May 4th 2010 at 04:30 PM.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    For an equation of nth degree given by

    ax^n +b* x^{n-1} + c*x^{n-2}..... = 0

    The sum of all its roots

    x_1 + x_2 + x_3 ...+x_n = -b/a

    --------------------------------

    I hope you know the equation of normal passing through a point on parabola.

    In that third degree equation

    x^2 term will be missing... this tells you that

    the value of its coefficient is 0

    Hence the sum of roots= (X1+X2+X3) = 0 /a = 0
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  3. #3
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    Quote Originally Posted by Blasnmt23 View Post
    Hello!
    I was assigned this for my Calc class and was wondering if anyone could give me some help in figuring this out. If you need any more information I can try and get it to you. I can upload the picture given if it helps also.

    Suppose that three points on the parabola y = x^2 have the property that their normal lines intersect at a common point. Show that the sum of their x-coordinates is zero. Show that x1 + x2 + x3 = 0.
    The derivative of y= x^2 is y'= 2x. In particular, the derivative at (x_0, x_0^2) is 2x_0 and so the equation of the normal line is y= -\frac{1}{2x_0}(x- x_0)+ x_0^2. The three normal lines, then, are
    y= -\frac{1}{2x_0}(x- x_0)+ x_0^2
    y= -\frac{1}{2x_1}(x- x_1)+ x_1^2
    and
    y= -\frac{1}{2x_2}(x- x_2)+ x_2^2

    Saying that the all intersect at a common point means there exist a specific x, say \overline{x} such that
    -\frac{1}{2x_0}(\overline{x}- x_0)+ x_0^2= -\frac{1}{2x_1}(\overline{x}- x_1)+ x_1^2 and
    -\frac{1}{2x_0}(\overline{x}- x_0)+ x_0^2= -\frac{1}{2x_2}(\overline{x}- x_2)+ x_2^2

    Find the conditions on x_0, x_1, and x_2 so that those two equations have a common solution.
    Last edited by HallsofIvy; April 30th 2010 at 03:16 AM.
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  4. #4
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    Post

    ..
    Last edited by Blasnmt23; May 4th 2010 at 04:29 PM.
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  5. #5
    Like a stone-audioslave ADARSH's Avatar
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    I will make myself clearer...its another way of doing your problem.
    -----
    For a parabola

    x^2 = 4ay


    the parametric equation of its normal at any general point (2at,at^2) is

    (y-at^2) = \frac{-1}{4at} (x-2at)

    (-4ayt+4a^2t^3) =  (x-2at)

    -4ayt+4a^2t^3 -x+2at = 0

    4a^2t^3~+~ 0*t^2 ~+(2-4y)at~-x = 0

    This is a third degree equation....
    Three different values of t satisfy the above equation ....
    This means normals at these three different t intersect at (x,y).

    (This can further be interpreted as your question )

    Now the sum of three roots is zero .
    ie;
    t_1 + t_2 +t_3 = 0 (as given by first post of mine)

    This implies

    2at_1 + 2at_2 + 2at_3 = 0

    But 2at_1 = x_1 and so on

    Thus

    x_1 + x_2 + x_3 = 0

    In your question it is to be proved for a = 1/4 but here we have proved it for all a.

    ----------
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  6. #6
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    I love math done beautifully...Thank you very much!
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