Explanation

(a)
General point on the curve $y^{2} = x - 1$ is $\left( t^{2} + 1,\ t \right)$ and the general point on the curve $x^{2} = y - 1$ is $(t,\ t^{2} + 1)$. Since, both curves are symmetrical about line $y = x$. For nearest point on curve $y^{2} = x - 1$ from the line $y = x$
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Let $D = \frac{t^{2} + 1 - t}{\sqrt{2}}$
$\Rightarrow \frac{dD}{dt} = \frac{1}{\sqrt{2}}(2t - 1)$
Put $\frac{dD}{dt} = 0 \Rightarrow t = \frac{1}{2}$. Then point is $\left( \frac{5}{4},\ \frac{1}{2} \right)$
Similarly, point on the other curve is $\left( \frac{1}{2},\ \frac{5}{4} \right)$
Distance between them
$= \sqrt{\left( \frac{5}{4} - \frac{1}{2} \right)^{2} + \left( \frac{1}{2} - \frac{5}{4} \right)^{2}}$
$= \sqrt{\frac{18}{16}} = \frac{3\sqrt{2}}{4}$