<s>[INST] 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!
 [/INST] $b^2a+a^2b=c^2a+a^2c=c^2b+b^2c\rightarrow b^2a+a^2b=c^2a+a^2c$
$\rightarrow b^2a-c^2a=-a^2b+a^2c\rightarrow (b^2-c^2)a=-a^2(b-c)$
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