Show that the sum of the $x$-coordinates of three points on the graph of $y = x^2$ whose normal lines intersect at a common point is $0$. 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've done a bit of work in trying to explain this, but I'm not so certain I can visualise it to begin with.

The the line tangent to the curve $y = x^2$ at a point $T(a, a^2)$ is $$y - a^2 = 2a(x - a) \implies y = a^2 + 2ax - 2a^2 \implies y = 2ax - a^2$$
In order for the lines tangent to two points to intersect at a common point $P(x, y)$, which will be along the $y$-axis, their $y$-intercepts must be equivalent, and $x = 0$.
The two points must therefore be equidistant from the origin, such $|c| = |a|$ in order for the lines tangent to their coordinates $T_2(c, c^2)$ and $T_1(a, a^2)$ will intersect at point $(x, y)$, as their $y$-intercepts are expressed by $-c^2$ and $-a^2$, respectively.

I feel a bit as though I'm rambling, to a certain extent. I'm having trouble thinking of a way to describe this scenario as it would seem that the normal line to each point would have to pass through the origin in order for the lines to meet at one point, as a linear function $f(x)$ may only meet another linear function $g(x)$ at one point assuming that $f(x) \neq g(x)$ and $m_f \neq m_g$ (which would mean infinitely many points of intersection, and no points of intersection, respectively).

Edit: 
Let there be three points $A(a,a^2)$, $B(b,b^2)$, $C(c,c^2)$ whose normal lines intersect at a point $P(x_p, y_p)$
The tangent to the curve at $A$ is $y - a^2 = 2a(x - a)$
The normal to the curve at $A$ is $y - a^2 = -\frac{1}{2a}(x - a) \implies x+2ay=2a^3+a$ 
$$
\begin{align*}
 \\ &\left[y = \dfrac{2a^3 + a - x}{2a} \right] \text{ } \left[y = \dfrac{2b^3 + b - x}{2b} \right] \text{ } \left[y = \dfrac{2c^3 + c - x}{2c} \right]
 \\
 \\ &\left[x = 2a^3 + a - 2ay \right] \left[ x = 2b^3 + b - 2by \right] \left[ x = 2c^3 + c - 2cy \right]
\end{align*}
$$
$I_{A \cdot B} = $ the intersection of $line_{normal_A}$ and $line_{normal_B} \implies$ $I_{A \cdot B}([-2ab(a+b)], [a^2+ab+b^2+\frac{1}{2}])$
$I_{A \cdot C} = $ the intersection of $line_{normal_A}$ and $line_{normal_C} \implies$ $I_{A \cdot C}([-2ac(a+c)], [a^2+ac+c^2+\frac{1}{2}])$

From here, I'm not certain how to proceed.
 A: Assume that $P_1,P_2$ are points on the parabola whose normal lines meet at $V$.
Then the circle $\Gamma_1$ with center $V$ through $P_1$ is tangent to the parabola, as well as the circle $\Gamma_2$ with center $V$ through $P_2$. If the normal in $P_3$ goes through $V$, too, then we have three concentric circles tangent to the same parabola, or, given $V=(v_x,v_y)$, three stationary points for the squared distance from $V$, i.e.:
$$ f(x) = (v_x-x)^2+(v_y-x^2)^2 .$$
That implies that $f'(x)$ has three real roots, in the $x$-coordinates of $P_1,P_2,P_3$. Since:
$$ f'(x) = 4x^3-(4v_y-2)x-2v_x, $$
the sum of the $x$-coordinates of $P_1,P_2,P_3$ is zero by Viète's theorem (the sum of the roots of $f'(x)$ is given by the opposite of the coefficient of $x^2$, i.e. $0$).
$$\phantom{}$$

Update: Since the normal line through $(x_0,x_0^2)$ intersects the $x$-axis in the point $(x_0+2x_0^3,0)$, the concurrence of the normal lines implies:
$$\det\left(\begin{array}{ccc}1&x_1&x_1^3 \\ 1&x_2&x_2^3 \\ 1&x_3&x_3^3\end{array}\right)=0$$
that implies $x_1+x_2+x_3=0$.
A: 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 $0$.
Let the points be $A(a,a^2)$, $B(b,b^2)$, $C(c,c^2)$
The tangent to the curve at $A$ is $y - a^2 = 2a(x - a)$
The normal to the curve at $A$ is $y - a^2 = -\frac{1}{2a}(x - a) \implies x+2ay=2a^3+a$ 
\begin{align*}
\\ &\left[y = \dfrac{2a^3 + a - x}{2a} \right] \text{ } \left[y = \dfrac{2b^3 + b - x}{2b} \right] \text{ } \left[y = \dfrac{2c^3 + c - x}{2c} \right]
\\ &\left[x = 2a^3 + a - 2ay \right] \left[ x = 2b^3 + b - 2by \right] \left[ x = 2c^3 + c - 2cy \right]
\end{align*}
$I_{A \cdot B} = $ intersection of the normal lines of $A$ and $B$ $\implies$ $I_{A \cdot B}([-2ab(a+b)], [a^2+ab+b^2+\frac{1}{2}])$
$I_{A \cdot C} = $ intersection of the normal lines of $A$ and $C$ $\implies$  $I_{A \cdot C}([-2ac(a+c)], [a^2+ac+c^2+\frac{1}{2}])$
$$
\\ \begin{align}
\\ I_{A \cdot B} = I_{A \cdot C} \implies x_{I_{A \cdot B}} = x_{I_{A \cdot C}} \implies \frac{x_{I_{A \cdot B}}}{x_{I_{A \cdot C}}} = 1 \implies \dfrac{-2ab(a+b)}{-2ac(a+c)} &= 1
\\
\\ b(a + b) &= c(a + c)
\\ ba+b^2 &= ca + c^2 
\\ a(b - c) &= c^2 - b^2
\\ a(b - c) &= (c + b)(c - b)
\\ a(b-c) &= (c+b)(b-c)(-1) 
\\  \frac{a(b-c)}{(b-c)} &= -c - b
\\ a + b + c &= 0
\\
\\\end{align}
$$
This shows the sum of $a+b+c$ is $0$.
