Results 1 to 6 of 6

Thread: Normal lines on a parabola

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    4

    Normal lines

    Thanks!
    Last edited by Blasnmt23; May 4th 2010 at 04:30 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    For an equation of nth degree given by

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

    The sum of all its roots

    $\displaystyle x_1 + x_2 + x_3 ...+x_n = -b/a$

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

    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,725
    Thanks
    3008
    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 $\displaystyle y= x^2$ is y'= 2x. In particular, the derivative at $\displaystyle (x_0, x_0^2)$ is $\displaystyle 2x_0$ and so the equation of the normal line is $\displaystyle y= -\frac{1}{2x_0}(x- x_0)+ x_0^2$. The three normal lines, then, are
    $\displaystyle y= -\frac{1}{2x_0}(x- x_0)+ x_0^2$
    $\displaystyle y= -\frac{1}{2x_1}(x- x_1)+ x_1^2$
    and
    $\displaystyle y= -\frac{1}{2x_2}(x- x_2)+ x_2^2$

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

    Find the conditions on $\displaystyle x_0$, $\displaystyle x_1$, and $\displaystyle x_2$ so that those two equations have a common solution.
    Last edited by HallsofIvy; Apr 30th 2010 at 03:16 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2010
    Posts
    4

    Post

    ..
    Last edited by Blasnmt23; May 4th 2010 at 04:29 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    I will make myself clearer...its another way of doing your problem.
    -----
    For a parabola

    $\displaystyle x^2 = 4ay$


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

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

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

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

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

    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;
    $\displaystyle t_1 + t_2 +t_3 = 0$ (as given by first post of mine)

    This implies

    $\displaystyle 2at_1 + 2at_2 + 2at_3 = 0 $

    But $\displaystyle 2at_1 = x_1$ and so on

    Thus

    $\displaystyle x_1 + x_2 + x_3 = 0 $

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

    ----------
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2010
    Posts
    4
    I love math done beautifully...Thank you very much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. area between parabola and lines SOS
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Dec 22nd 2011, 02:37 AM
  2. Parabola and normal lines
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Apr 23rd 2010, 03:52 PM
  3. Replies: 6
    Last Post: Oct 18th 2009, 05:50 PM
  4. Replies: 2
    Last Post: Jan 25th 2009, 08:41 PM
  5. Lines Tangent to a Parabola
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 22nd 2008, 04:08 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum