Normal lines on a parabola

**Blasnmt23** · April 28th 2010, 06:42 PM

Thanks!

**ADARSH** · April 28th 2010, 10:01 PM

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

**HallsofIvy** · April 29th 2010, 03:24 AM

Originally Posted by Blasnmt23

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.

**Blasnmt23** · April 29th 2010, 11:32 AM

..

**ADARSH** · April 29th 2010, 10:12 PM

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.

----------

**Blasnmt23** · April 29th 2010, 10:26 PM

I love math done beautifully...Thank you very much!

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