To remove the x and y terms of the equation $14x^2–4xy+11y^2–36x+48y+41=0$ the shifted origin is

$$\begin{gathered} \mathrm{Given} \mathrm{equation} 14{\mathrm{x}}^2-4\mathrm{xy}+11{\mathrm{y}}^2-36\mathrm{x}+48\mathrm{y}+41=0 \\ \mathrm{compare} \mathrm{with} {\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{gy}+\mathrm{c}=0 \\ \mathrm{then} \mathrm{a}=14, 2\mathrm{h}=-4, \mathrm{b}=11, 2\mathrm{g}=-36, 2\mathrm{f}=48, \mathrm{c}=41 \\ \Rightarrow \mathrm{a}=14, \mathrm{h}=-2, \mathrm{b}=11, \mathrm{g}=-18, \mathrm{f}=24, \mathrm{c}=41 \\ \mathrm{Required} \mathrm{point} =\left[\frac{\mathrm{hf}-\mathrm{bg}}{\mathrm{ab}-{\mathrm{b}}^2},\frac{\mathrm{gh}-\mathrm{af}}{\mathrm{ab}-{\mathrm{b}}^2}\right] \\ =\left[\frac{(-2)\left(24\right)-11(-18)}{14\left(11\right)-{(-2)}^2},\frac{(-18)(-2)-14\left(24\right)}{14\left(11\right)-{(-2)}^2}\right] \\ =\left[\frac{-48+198}{154-4},\frac{36-336}{154-4}\right] \\ =\left[\frac{150}{150},\frac{-300}{150}\right] \\ =\left[1,-2\right] \end{gathered}$$