1. **State the problem:** Find the minimum distance from the point $P(4,2)$ to the parabola given by the equation $$y^2 = 8x.$$\n\n2. **Formula for distance:** The distance $d$ from a point $(x_0,y_0)$ to a point $(x,y)$ on the curve is given by $$d = \sqrt{(x - x_0)^2 + (y - y_0)^2}.$$\n\n3. **Set up the distance squared function:** To minimize distance, we minimize the squared distance to avoid the square root: $$D = (x - 4)^2 + (y - 2)^2.$$\n\n4. **Use the parabola equation:** Since $y^2 = 8x$, express $x$ in terms of $y$: $$x = \frac{y^2}{8}.$$\n\n5. **Rewrite $D$ in terms of $y$ only:** $$D(y) = \left(\frac{y^2}{8} - 4\right)^2 + (y - 2)^2.$$\n\n6. **Expand and simplify:**\n$$\left(\frac{y^2}{8} - 4\right)^2 = \left(\frac{y^2}{8}\right)^2 - 2 \cdot 4 \cdot \frac{y^2}{8} + 16 = \frac{y^4}{64} - y^2 + 16,$$\n$$D(y) = \frac{y^4}{64} - y^2 + 16 + y^2 - 4y + 4 = \frac{y^4}{64} - 4y + 20.$$\n\n7. **Find critical points by differentiating $D(y)$:**\n$$D'(y) = \frac{4y^3}{64} - 4 = \frac{y^3}{16} - 4.$$\n\n8. **Set derivative to zero to find minima:**\n$$\frac{y^3}{16} - 4 = 0 \implies y^3 = 64 \implies y = 4.$$\n\n9. **Find corresponding $x$:**\n$$x = \frac{y^2}{8} = \frac{16}{8} = 2.$$\n\n10. **Calculate minimum distance:**\n$$d = \sqrt{(2 - 4)^2 + (4 - 2)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.828.$$\n\n**Answer:** The minimum distance is approximately $2.828$, which corresponds to option D (2222).