Suppose that three points on the parabola $y=x^2$ have property that their normal lines intersect at a common point (m,n) . We have to find The sum of their x coordinates .
I just know equation of normal with three different slope woukd be $y=mx-2am-am^3$ .
Here a=1/4 and m is slope of normal .
but now how to proceed .
Suppose that the normal at a point $(h,h^2)$ on the parabola passes through $(m,n)$. Then $$-1=\left(\left.\frac{dy}{dx}\right|_{x=h}\right)\left(\frac{h^2-n}{h-m}\right)=2h\left(\frac{h^2-n}{h-m}\right)$$ Therefore, $-(h-m)=2h^3-2nh$ and hence $2h^3-(2n-1)h-m=0$.
The $x$-coordinates of the three points are the three roots of the above cubic equation in $h$. So the sum of the $x$-coordinates is $0$.