General point on the curve $y^2=x-1$ is $\left(t_1^2+1,t_1\right)$ and the general point on the curve $x^2=y-1$ is $\left(t_2,t_2^2+1\right).$ Since both the curves are symmetrical about the line $y=x$, for the nearest point on the curve $y^2=x-1$ from the line $y=x$

Let, $Q\left(t_1^2+\mathrm{1,} t_1 \right)$

$P(t_2,t_2^2+1)$

slopes of the tangents are same

$\frac{1}{2t_1}=2\cdot t_2$

$t_1{t}_2=\frac{1}{4}$......(1)

Since $PQ$ is perpendicular to line $x=y$

$\frac{t_1-t_2^2-1}{t_1^2+1-t_2}=-1$......(2)

Solving (1) & (2),  we get $t_1=t_2=\frac{1}{2}$

$P\left(\frac{1}{2},\frac{5}{4}\right), Q\left(\frac{5}{4},\frac{1}{2}\right)$

therefore, $PQ=\frac{3\sqrt{2}}{4}$