Explanation

(a)
Given equation of parabola is $y = x^{2}$ ...(i)
Equation of straight line is $y = 2x - 4$ ...(ii)
On solving Eqs. (i) and (ii), we get
$x^{2} - 2x + 4 = 0$
Let $\mathcal{z} = x^{2} - 2x + 4$
$\therefore\mathcal{z}' = 2x - 2$
For least value, $\mathcal{z}^{'} = 0 \Rightarrow 2x - 2 = 0 \Rightarrow x = 1$
$\mathcal{z}''$ is positive at $x = 1$
$\therefore$ It is minimum, putting $x = 1$ in Eq. (i), we get
$y = 1$
So, the required point at the least distance from the line is $(1,\ 1)$