1. **State the problem:** We need to find the area of the region enclosed by the curve $y^2 = 4ax$ and the line $x = a$, where $a > 0$. 2. **Understand the curve:** The curve $y^2 = 4ax$ is a parabola opening to the right. 3. **Find the points of intersection:** The line $x = a$ intersects the parabola at points where $y^2 = 4a \cdot a = 4a^2$, so $y = \pm 2a$. 4. **Set up the integral for the area:** The area enclosed is between $x=0$ (vertex of parabola) and $x=a$. For each $x$, $y$ ranges from $-2\sqrt{ax}$ to $2\sqrt{ax}$. The area is $$\text{Area} = \int_0^a (\text{top } y - \text{bottom } y) \, dx = \int_0^a 2\sqrt{4ax} \, dx = \int_0^a 2 \cdot 2\sqrt{ax} \, dx = \int_0^a 4\sqrt{ax} \, dx.$$ 5. **Simplify the integral:** Since $\sqrt{ax} = \sqrt{a} \sqrt{x}$, $$\text{Area} = 4\sqrt{a} \int_0^a \sqrt{x} \, dx = 4\sqrt{a} \int_0^a x^{1/2} \, dx.$$ 6. **Evaluate the integral:** $$\int_0^a x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^a = \frac{2}{3} a^{3/2}.$$ 7. **Substitute back:** $$\text{Area} = 4\sqrt{a} \cdot \frac{2}{3} a^{3/2} = \frac{8}{3} \sqrt{a} \cdot a^{3/2} = \frac{8}{3} a^{(1/2 + 3/2)} = \frac{8}{3} a^2.$$ 8. **Conclusion:** The area is $\frac{8}{3} a^2$, so $k = \frac{8}{3}$. **Final answer:** $k = \frac{8}{3}$.