Three Distinct Points and Their Normal Lines 
Suppose That three points on the graph of $y=x^2$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$-coordinates is $0$.

I have a lot going but can not finish it.
Proof:
Let $(a,a^2)$, $(b,b^2)$, and $(c,c^2)$ be three distinct points on $y=x^2$ such that $a\not=b\not=c$.
Find the tangent slope and the slope of their normal lines.
$$(a,a^2) \hspace{4mm}m_{tan}=2a, \hspace{4mm} m_{norm}=\frac{-1}{2a}$$
$$(b,b^2) \hspace{4mm}m_{tan}=2b, \hspace{4mm} m_{norm}=\frac{-1}{2b}$$
$$(c,c^2) \hspace{4mm}m_{tan}=2c, \hspace{4mm} m_{norm}=\frac{-1}{2c}$$
Normal Line $(a,a^2)$
$y-a^2=-\frac{1}{2a}(x-a) \implies y=-\frac{1}{2a}x+\frac{1}{2}+a^2$
Normal Line $(b,b^2)$
$y-b^2=-\frac{1}{2b}(x-b) \implies y=-\frac{1}{2b}x+\frac{1}{2}+b^2$
Normal Line $(c,c^2)$
$y-c^2=-\frac{1}{2c}(x-c) \implies y=-\frac{1}{2c}x+\frac{1}{2}+c^2$
Find the $x$-interception points between the normal lines of $(a,a^2)$ and $(b,b^2)$.
$-\frac{1}{2a}x+\frac{1}{2}+a^2=-\frac{1}{2b}x+\frac{1}{2}+b^2 \implies x=-(b+a)2ab$
Find the $x$-interception points between the normal lines of $(a,a^2)$ and $(c,c^2)$.
$-\frac{1}{2a}x+\frac{1}{2}+a^2=-\frac{1}{2c}x+\frac{1}{2}+c^2 \implies x=-(c+a)2ac$
Find the $x$-interception points between the normal lines of $(b,b^2)$ and $(c,c^2)$.
$-\frac{1}{2b}x+\frac{1}{2}+b^2=-\frac{1}{2c}x+\frac{1}{2}+c^2 \implies x=-(c+b)2bc$
Show that $a+b+c=0$
$$\begin{align}
....\\
....\\
....\\
\end{align}$$
I do not know how to show that $a+b+c=0$. Any advice on how to continue? Thanks in advance!
 A: Since all normal lines intersect at a common point you have
$$
-(b+a)2ab=-(c+a)2ac=-(c+b)2bc,
$$
which means
$$
b^2a+a^2b=c^2a+a^2c=c^2b+b^2c.
$$
Now if $a=0$
$$
c^2b+b^2c=0\Rightarrow c=-b.
$$
If $a\neq0$,since $b\neq c$, from the first equality
$$
(b^2-c^2)a=-a^2(b-c) \Rightarrow b+c=-a.
$$
A: Michele Maschio’s answer picks up nicely where you left off, although neither your partial solution nor that answer deals with the case that one of the points is the origin, where the slope of the normal is undefined. This is a slightly different approach that doesn’t need to consider that case separately.  
If $y=f(x)$ is differentiable at $x=x_0$, then an equation of the line normal to the curve at the point $(x_0,f(x_0))$ is $(x-x_0)+f'(x_0)(y-f(x_0))=0$. In contrast to the slope-intercept form of the equation for a line, this form is also valid when $f'(x_0)=0$. For this problem the equation is $$(x-x_0)+2x_0(y-x_0^2)=0$$ or, after rearranging a bit, $$x+2x_0y=x_0(1+2x_0^2).$$ The condition that the normals through three distinct points with $x$-coordinates $a$, $b$ and $c$ gives us the system $$\begin{align}x+2ay&=a(1+2a^2)\\x+2by&=b(1+2b^2)\\x+2cy&=c(1+2c^2).\end{align}$$ Subtract the first equation from the other two to eliminate $x$: $$\begin{align}2(b-a)y&=2b^3-2a^2+b-a\\2(c-a)y&=2c^3-2a^3+c-a.\end{align}$$ Cross-multiply and subtract to eliminate $y$, producing $$\begin{align}0&=(c-a)(2b^3-2a^3+b-a)-(b-a)(2c^3-2a^3+c-a) \\ &=2a^3b-2ab^3+2b^3c-2bc^3+2c^3a-2ca^3 \\ &= 2(b-a)(c-b)(a-c)(a+b+c).\end{align}$$ The last two expressions have a satisfying symmetry that reflects the symmetry of the roles of the three points in the problem. Since the three points are distinct, we must have $a+b+c=0$ for our system of equations to be consistent.
A: $b^2a+a^2b=c^2a+a^2c=c^2b+b^2c\rightarrow b^2a+a^2b=c^2a+a^2c$
$\rightarrow b^2a-c^2a=-a^2b+a^2c\rightarrow (b^2-c^2)a=-a^2(b-c)$
