The shortest distance between the line $\mathrm{y}-\mathrm{x}=1$ and the curve $\mathrm{x}={\mathrm{y}}^2$ is

$$\begin{gathered} \mathrm{Given} \mathrm{curve} \mathrm{x}={\mathrm{y}}^2 \\ \mathrm{y}-\mathrm{x}=1 \Rightarrow \mathrm{x}-\mathrm{y}+1=0 \\ \mathrm{let} \mathrm{P}\left({\mathrm{p}}^2,\mathrm{p}\right) \mathrm{be} \mathrm{a} \mathrm{point} \mathrm{on} \mathrm{the} \mathrm{parabola} \end{gathered}$$

Let d = perpendicular distance from P to line x-y+1=0

d = $\frac{\left|{\mathrm{p}}^2-\mathrm{p}+1\right|}{\sqrt{2}}$

$$\begin{gathered} \text{⨍}\left(\mathrm{p}\right){\text{=p}}^2\text{-p+1} \\ {\text{⨍}}^1\text{(p)=2p-1=0} \\ \Rightarrow \mathrm{p}=\frac{1}{2} \end{gathered}$$

point is $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

distance = $\frac{\left|{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+1\right|}{\sqrt{2}}$

          $=\frac{\left|{\displaystyle \frac{-1}{4}}+1\right|}{\sqrt{2}}=\frac{3}{4\sqrt{2}}=\frac{3\sqrt{2}}{8}$