For what values of a and b is the line 2x + y = b tangent to the parabola y = ax2 when x = 2?

Didn't you mean    y = ax^2?  "^" denotes "exponentiation."

The first derivative of y = ax^2 represents the slope of the tangent line to the curve of y = ax^2.  Here, dy/dx = 2ax.  When x = 2, dy/dx = 2a(2) = 4a.

The point of tangency is (2,y), where y = a(2)^2, or y=4a; thus, the point of tangency is (2,4a).  The equation of the tangent line to y=ax^2 at (2,4a) is found by (1) differentiating y=ax^2 with respect to x, (2) letting x = 2 in the result:        dy/dx = 2ax    =>    dy/dx (at 2,4a) = 2a(2) = 4a

The line 2x + y = b is supposed to be tangent to y = ax^2 at (2,4a).

The slope of 2x + y = b is found by solving 2x + y = b for y:

                             y = b - 2x        => slope m = -2

Thus, dy/dx = 4a = - 2, and thus a = -2/4, or a = -1/2.  All we have to do now is to find the value of b.   We know that 2x + y = b, so if x=-2 and y=-8, 

2(-2) + [-8] = b = -4 - 8 = -12

Thus, the equation of the parabola is   y = ax^2 = (-1/2)x^2.

a = -2 and b = -8 are the required a and b values.