When the origin is shifted to the point (2, 3) the transformed equation of a curve is ${\mathrm{x}}^2+3\mathrm{xy}-2{\mathrm{y}}^2+17\mathrm{x}-7\mathrm{y}-11=0$. Find the original equation of the curve.

Let (X, Y) be the new coordinates of the point (x, y), given (h, k) = (2, 3)
Given transformed equation is
$\begin{array}{l}{\mathrm{X}}^2+3\mathrm{XY}-2{\mathrm{Y}}^2+17\mathrm{X}-7\mathrm{Y}-11=0\\ \mathrm{Now} \mathrm{x}=\mathrm{X}+h,\mathrm{y}=\mathrm{Y}+\mathrm{k}\\ \Rightarrow \mathrm{X}=\mathrm{x}-\mathrm{h},\mathrm{Y}=\mathrm{y}-\mathrm{k}\\ \Rightarrow \mathrm{X}=\mathrm{x}-2,\mathrm{Y}=\mathrm{y}-3\end{array}$
$\therefore \text{ the original equation is }$
$\begin{array}{l}(\mathrm{x}-2{)}^2+3(\mathrm{x}-2\left)\right(\mathrm{y}-3)-2(\mathrm{y}-3{)}^2+17(\mathrm{x}-2)-7(\mathrm{y}-3)-11=0\\ \\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}+4+3\mathrm{xy}-9\mathrm{x}-6\mathrm{y}+18-2{\mathrm{y}}^2+12\mathrm{y}-18+17\mathrm{x}-34-7\mathrm{y}+21-11=0\\ \\ \Rightarrow {\mathrm{x}}^2+3\mathrm{xy}-2{\mathrm{y}}^2+4\mathrm{x}-\mathrm{y}-20=0\end{array}$