Thanks!

Printable View

- Apr 28th 2010, 06:42 PMBlasnmt23Normal lines
Thanks!

- Apr 28th 2010, 10:01 PMADARSH
For an equation of nth degree given by

The sum of all its roots

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

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 - Apr 29th 2010, 03:24 AMHallsofIvy
The derivative of is y'= 2x. In particular, the derivative at is and so the equation of the

**normal**line is . The three normal lines, then, are

and

Saying that the all intersect at a common point means there exist a specific x, say such that

and

Find the conditions on , , and so that those two equations have a common solution. - Apr 29th 2010, 11:32 AMBlasnmt23
..

- Apr 29th 2010, 10:12 PMADARSH
I will make myself clearer...its another way of doing your problem.

-----

For a parabola

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

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;

(as given by first post of mine)

This implies

But and so on

Thus

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

---------- - Apr 29th 2010, 10:26 PMBlasnmt23
I love math done beautifully...Thank you very much!