The equation of the tangent to the curve $\mathrm{y}=\frac{2}{\sqrt{\mathrm{x}-1}}$at the point $\mathrm{x}=10$ is

$\mathrm{x}=10 \mathrm{y}=\frac{2}{3}$

$\therefore$ Point$=\left(10, \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$

$\begin{array}{rcl}\frac{\mathrm{dy}}{\mathrm{dx}}& =& 2·\frac{-1}{2}{(\mathrm{x}-1)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ & =& -{(10-1)}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{-1}{27}\end{array}$

eq of Tangent

$\begin{array}{rcl}\mathrm{y}-\frac{2}{3}& =& \frac{-1}{27} (\mathrm{x}-10)\\ 27\mathrm{y}-18& =& -\mathrm{x}+10 \Rightarrow \mathrm{x}+27\mathrm{y}=28\end{array}$