Thanks!
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
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.
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.
----------