Answer: C

Small Hint:

Write $A = (a, a^2)$ and $B = (-a, a^2);$ since $A, B, C$ have distinct $x$-coordinates, the right angle is at $C$

Big Hint:

Perpendicular legs $CA$ and $CB$ force $a^2 - c^2 = 1,$ the height above $\overline{AB};$ get $|a|$ from the area, then $c^2$

Solution:

Since $AB$ is horizontal, take $A = (a, a^2)$ and $B = (-a, a^2),$ and let $C = (c, c^2).$ The right angle cannot be at $A$ or $B$ (that would need $c = \pm a$), so it is at $C.$

Put $d = a^2-c^2.$ Taking the dot product of $\overrightarrow{CA}$ and $\overrightarrow{CB}$ gives $d(d-1)=0.$ The points are distinct, so $d \ne 0;$ hence $a^2-c^2=1.$ This value is the height of the triangle above $\overline{AB}.$

The area is $\tfrac12 \cdot AB \cdot \text{height}$ $= \tfrac12 (2|a|)(1)$ $= |a| = 2008,$ so $a^2 = 2008^2 = 4{,}032{,}064$ and the $y$-coordinate of $C$ is $c^2 = a^2 - 1 = 4{,}032{,}063.$

Its digit sum is $4 + 0 + 3 + 2 + 0 + 6 + 3 = 18.$

Thus, the correct answer is C.