jmc

If these two graphs intersect then the point of intersection occur when $$x^2+a=ax,$$ or $$x^2-ax+a=0.$$ This quadratic has one solution exactly when the discriminant is equal to zero: $$(-a{)}^2-4\cdot 1\cdot a=0.$$ This simplifies to $$a(a-4)=0.$$

There are exactly two values of $a$ for which the line and parabola intersect one time, namely $a=0$ and $a=4$. The sum of these values is $$0+4=\boxed{4}.$$