Given:
y^2 = 8x (Equation of Parabola)
y = 2x (Equation of Line)

To find:
Area enclosed by the parabola and the line

Solution:
Let's plot the graph of the given equations.

We can see that the parabola and the line intersect at (2,4).

Let's find the points of intersection of the parabola and the line.

y^2 = 8x
Substituting y = 2x
(2x)^2 = 8x
4x^2 - 8x = 0
4x(x - 2) = 0
x = 0 or x = 2
When x = 0, y = 0
When x = 2, y = 4

So, the points of intersection are (0,0) and (2,4).

Let's find the equation of the tangent to the parabola at (2,4).

Differentiating y^2 = 8x
2y dy/dx = 8
dy/dx = 4/y
At (2,4), dy/dx = 4/4 = 1
Equation of tangent at (2,4) is y - 4 = 1(x - 2)
y = x + 2

Let's find the x-coordinate of the point where the tangent intersects the line.

Substituting y = x + 2 in y = 2x
x + 2 = 2x
x = 2

So, the point of intersection of the tangent and the line is (2,4).

Hence, the area enclosed by the parabola and the line is the area of the region bounded by the x-axis, the line y = 2x, the tangent to the parabola at (2,4), and the vertical line x = 0.

Let's find the equation of the parabola in terms of y.

y^2 = 8x
x = y^2/8
Substituting in y = 2x
y = 2(y^2/8)
y = y^2/4
y^2 - 4y = 0
y(y - 4) = 0
y = 0 or y = 4

So, the y-coordinates of the points of intersection of the parabola and the line are 0 and 4.

Therefore, the area enclosed by the parabola and the line is given by the integral of the difference between the line and the parabola from x = 0 to x = 2.

Area = ∫(2x - y^2/8)dx from x = 0 to x = 2
= ∫(2x - x^2/2)dx from x = 0 to x = 2 (Substituting y^2 = 8x)
= [x^2 - (x^3)/6] from x = 0 to x = 2
= (4/3)

Therefore, the area enclosed by the parabola and the line is 4/3.

Answer: (a) 4/3