The normal to the curve xy = 4 at the point (1,4) meets the curve again at the point

$$\begin{gathered} \begin{array}{r}\mathrm{xy}=4 \\ \Rightarrow \mathrm{y}=\frac{4}{\mathrm{x}}\Rightarrow \frac{d\mathrm{y}}{d\mathrm{x}}=\frac{-4}{{\mathrm{x}}^2}=-4 \mathrm{at} \mathrm{x}=1 \\ \mathrm{equation} \mathrm{of} \mathrm{normal} \mathrm{is} \mathrm{y}-4=\frac{1}{4}(\mathrm{x}-1)\\ 4\mathrm{y}-16=\mathrm{x}-1\end{array} \end{gathered}$$
$$\begin{gathered} Solving with xy = 4 \\ \Rightarrow x=-16,y=\frac{-1}{4} \end{gathered}$$