We can solve this problem by finding the point of intersection between the line and the parabola.

First, we rearrange the equation of the line x-y 2=0 into slope-intercept form:

y = x - 2

Next, we substitute this equation into the equation of the parabola:

y^2 = 8x

(y becomes (x-2)^2)

(x-2)^2 = 8x

Expanding and rearranging, we get:

x^2 - 12x + 16 = 0

Using the quadratic formula, we can solve for x:

x = (12 +/- sqrt(144 - 64)) / 2

x = 8 or 4

Substituting these values of x back into the equation of the line, we can find the corresponding y values:

When x = 8: y = 6

When x = 4: y = 2

Therefore, the point of intersection between the line and the parabola is (8, 6) or (4, 2).

To determine which point corresponds to the point of tangency, we must check which point has a slope that is equal to the slope of the tangent line at that point.

The slope of the tangent line to the parabola y^2 = 8x at any point (x,y) is:

dy/dx = 4/y

At the point (2,-4), the slope of the tangent line is:

dy/dx = 4/-4 = -1

At the point (1,2), the slope of the tangent line is:

dy/dx = 4/2 = 2

Since the slope of the line x-y 2=0 is y = x - 2, which has a slope of 1, we see that the line is tangent to the parabola at the point (2,-4).

Therefore, the answer is (a) (2,-4).