Solution:

Let $A=\left(r, r^{2}\right)$ and $B=\left(s, s^{2}\right)$. Since $r$ and $s$ are roots of $x^{2}-a x-b$ with midpoint $5$, $r+s=10=a$ (where the last equality follows by Vieta's formula).

Now, as $-r s=b$ (Vieta's formula), observe that

$$202=r^{2}+s^{2}=(r+s)^{2}-2 r s=100+2 b.$$

This means $b=51$, so the answer is $10+51=61$.

Solution:

As in the previous solution, let $A=\left(r, r^{2}\right)$ and $B=\left(s, s^{2}\right)$ and note $r+s=10=a$.

Fixing $a=10$, the $y$-coordinate of the midpoint is $50$ when $b=0$ (and changing $b$ shifts the line up or down by its value). So, increasing $b$ by $51$ will make the midpoint have $y$-coordinate $50+51=101$, so the answer is $10+51=61$.