Given equation is $\mathrm{y}^{2}-8 \mathrm{x}$ Take an arbitrary point on this curve. If we take y as $\mathrm{P}$.
then point is $\left(\frac{P^{2}}{8}, P\right)$.
The distance between $\left(\frac{P^{2}}{8}, P\right)$ and $(4,2)$ is
$d^{2}=\left(\frac{P^{2}}{8}-4\right)^{2}+(P-2)^{2}$ ...(1)
$=\frac{1}{64}\left(P^{2}-32\right)^{2}+(P-2)^{2}$
$\Rightarrow 2 d \cdot \frac{d d}{d p}=\frac{1}{64} \times 2\left(P^{2}-32\right) \times 2 P+2(P-2)$
$=\frac{1}{16}\left(P^{2}-32\right) P+2(P-2)$
for minimum distance, $\frac{d d}{d p}=0$ $\Rightarrow \mathrm{P}^{3}-32 \mathrm{P}+32 \mathrm{P}-64=0$
$\Rightarrow \mathrm{P}^{3}=64$
$\Rightarrow \mathrm{P}=4$
$\therefore(1) \Rightarrow d^{2}=\left(\frac{16}{8}-4\right)^{2}+(4-2)^{2}$
$=(-2)^{2}+(2)^{2}=8$
$\Rightarrow d=\sqrt{8}=2 \sqrt{2}$