Point $(4,2)$ isn't on the curve

The shortest distance to $P(4,2)$ will be from the point $A\left(\dfrac{a^{2}}{8},a\right)$ on $y^{2} = 8x$ at which the normal to the tangent of the curve at $A$ passes through $P$ .

Now by implicit differentiation we have that $2yy' = 8 \Longrightarrow y' = \dfrac{4}{y}$ . Thus the slope of the normal to the tangent at $A$ will be $-\dfrac{y}{4}$ , and the equation of the normal will then be $y - a = -\dfrac{a}{4}\left(x - \dfrac{a^{2}}{8}\right)$ . Plugging in the coordinates of $P$ , (as we want $P$ to be on this line), we have that

$2 - a = -\dfrac{a}{4}\left(4 - \dfrac{a^{2}}{8}\right) \Longrightarrow 8 - 4a = -4a + \dfrac{a^{3}}{8} \Longrightarrow a^{3} = 64 \Longrightarrow a = 4, \dfrac{a^{2}}{8} = 2$ .

The closest point on the curve to $P(4,2)$ is thus $A(2,4)$ , and $|AP| = \sqrt{(4 - 2)^{2} + (2 - 4)^{2}} = \boxed{2\sqrt{2}}$ .

The distance between point $(4,2)$ and $y^2=8x$ is given by $\sqrt{(x-4)^2+(y-2)^2}$ . By Cauchy-Schwarz inequality , we have:

$\begin{aligned} \sqrt{(x-4)^2+(y-2)^2} & \ge \frac {|x-4|+|y-2|}{\sqrt 2} \end{aligned}$

Equality occurs when:

$\begin{aligned} |x-4| & = |y-2| \\ \left|\frac {y^2}8 - 4\right| & = |y - 2| \\ |y^2-32| & = 8|y-2| \\ \implies y & = 4 \\ \implies x & = \frac {4^2}8 = 2 \end{aligned}$

And the smallest distance is: $\dfrac {|x-4|+|y-2|}{\sqrt 2} = \dfrac {|2-4|+|4-2|}{\sqrt 2} = \boxed{2\sqrt 2}$