The minimum value of ${\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}^2+{\left(\frac{{\mathrm{x}}_1^2}{20}-\sqrt{\left(17-{\mathrm{x}}_2\right)\left({\mathrm{x}}_2-13\right)}\right)}^2$ , where ${\mathrm{x}}_1\in [13,17]$  , is

$$\begin{gathered} A\left({\mathrm{x}}_1,{\mathrm{y}}_1\right) lies on \mathrm{y}=\frac{{\mathrm{x}}^2}{20},\Rightarrow {\mathrm{x}}^2=20\mathrm{y} \\ B\left({\mathrm{x}}_2,{\mathrm{y}}_2\right) lies on \mathrm{y}=\sqrt{(17-\mathrm{x})(\mathrm{x}-13)}\Rightarrow y^2=17x-221-x^2+13x \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-30\mathrm{x}+221=0 \\ Equation of normal to {\mathrm{x}}^2=20\mathrm{y} \mathrm{at} A(10\mathrm{t},5{\mathrm{t}}^2) \\ \mathrm{is} \mathrm{x}+\mathrm{yt}=10\mathrm{t}+5{\mathrm{t}}^3 \\ \mathrm{This} \mathrm{normal} \mathrm{passes} \mathrm{through} \mathrm{centre} \mathrm{of} \mathrm{circle} (15,0) \\ 10\mathrm{t}+5{\mathrm{t}}^3=15\Rightarrow \mathrm{t}=1 \\ \mathrm{A}(10,5),\mathrm{C}(15,0),\mathrm{r}=2 \\ \mathrm{AB}=\mathrm{Ac}-\mathrm{r}=5\sqrt{2}-2 \\ \mathrm{Required} \mathrm{minimum} \mathrm{value} \mathrm{of}{\left( \mathrm{AB}\right)}^2={\left(5\sqrt{2}-2\right)}^2 \end{gathered}$$