Explanation

(b)
Shortest distance between two curves occured along the common normal
Normal to $y^{2} = 4x$ at $(m^{2},\ 2m)$ is
$y + mx - 2m - m^{3} = 0$
Normal to $y^{2} = 2(x - 3)$ at $\left( \frac{m^{2}}{2} + 3,\ m \right)$ is
$y + m(x - 3) - m - \frac{m^{3}}{2} = 0$
Both normals are same, if $- 2m - m^{3} = - 4m - \frac{1}{2}m^{3}$
$\Rightarrow m = 0,\ \pm 2$
So, points will be $(4,\ 4)$ and $(5,\ 2)$ or $(4,\ - 4)$ and $(5,\ - 2)$
Hence, shortest distance will be
$\sqrt{1 + 4} = \sqrt{5}$