First, note that the derivative of y=x^2 is y'=2x, which means that the slope of the tangent at (x,x^2) is 2x so the slope of the normal at (x,x^2) is -1/2x.
An equation of the normal at (a,a^2) is
y-a^2=-1/(2a)(x-a)
Observe that the line intersection y=x^2 at the solutions of the system
{y=x^2,y-a^2=-1/(2a) (x-a)}:
(a,a^2) and (-1/(2a)-a,1+1/(4a^2)+a^2).
We are trying to find a value of a that minimizes the distance between these two points. That is, we want to minimize
l(a)=(-1/(2a)-a-a)^2+
(1+1/(4a^2)+a^2-a^2)^2
=(1+4a^2)^3/(16a^4).
Differentiating gives us
l'(a)=(-1-6a^2+32a^6)/(4a^5)
=(2a^1-1)(4a^2+1)^2 / (4a^5)
and solving
l'(a)=0 gives us a=-1/\sqrt{2} or a=1/\sqrt{2}.
