Let $A$, $B$ and $C$ be three distinct points on the graph of $y=x^2$ such that line $AB$ is parallel to the $x$-axis and $\triangle ABC$ is a right triangle with area $2008$. What is the sum of the digits of the $y$-coordinate of $C$?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$
Supposing $\angle A=90^\circ$, $AC$ is perpendicular to $AB$ and, it follows, to the $x$-axis, making $AC$  a segment of the line $x=m$. But that would mean that the coordinates of $C$ are $(m, m^2)$, contradicting the given that points $A$ and $C$ are distinct. So $\angle A$ is not $90^\circ$. By a similar logic, neither is $\angle B$. 
This means that $\angle C=90^\circ$ and $AC$ is perpendicular to $BC$. Let C be the point $(n, n^2)$.  So the slope of $BC$ is the negative reciprocal of the slope of $AC$, yielding $m+n=\frac{1}{m-n}$ $\Rightarrow$ $m^2-n^2=1$. 
Because $m^2-n^2$ is the length of the altitude of triangle $ABC$ from $AB$, and $2m$ is the length of $AB$, the area of $\triangle ABC=m(m^2-n^2)=2008$. Since $m^2-n^2=1$, $m=2008$. 
Substituting, $2008^2-n^2=1$ $\Rightarrow$ $n^2=2008^2-1=(2000+8)^2-1=4000000+32000+64-1=4032063$, whose digits sum to $18 \Rightarrow \textbf{(C)}$.