Explanation

(a)
Given,
$y - x = 1$
$\Rightarrow \ \ \ \ \ y = x + 1$
$\frac{dy}{dx} = 1$
And $y^{2} = x$
$\Rightarrow \ \ \ \ \ \ 2y\frac{dy}{dx} = 1$
$\Rightarrow \ \ \ \ \ \frac{dy}{dx} = \frac{1}{2y}$
$\therefore\ \ \ \ \ 1 = \frac{1}{2y}$
$\Rightarrow \ \ \ \ \ 2y = 1$
$\Rightarrow \ \ \ \ y = \frac{1}{2}$
$\therefore\$point on the curve is $\left( \frac{1}{4},\frac{1}{2} \right)$
$\therefore\ \ \$required shortest distance
$= \left| \frac{\frac{1}{4} - \frac{1}{2} + 1}{\sqrt{2}} \right| = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8}$