If a hyperbola passes through the point P(10, 16) and it has vertices at $(\pm 6,0)$,then the equation of the normal at P is

$$\begin{gathered} \mathrm{let} \mathrm{hyperbola} \mathrm{be} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 — ① \\ \mathrm{Given} \mathrm{verties} (\pm \mathrm{a}, 0)=(\pm 6, 0) \\ \Rightarrow a=6 \\ Since ① \mathrm{passes} \mathrm{through} \mathrm{the} \mathrm{point} (10, 16) \mathrm{then} \\ \frac{100}{{\mathrm{a}}^2}-\frac{256}{{\mathrm{b}}^2}=1 \\ \Rightarrow \frac{100}{36}-\frac{256}{{\mathrm{b}}^2}=1 \\ \Rightarrow \frac{100}{36}-1=\frac{256}{{\mathrm{b}}^2} \\ \Rightarrow \frac{256}{{\mathrm{b}}^2}=\frac{64}{36} \\ \Rightarrow {\mathrm{b}}^2=\frac{256\times 36}{64}=144 \\ \therefore From ① hyperbola is \frac{{\mathrm{x}}^2}{36}-\frac{{\mathrm{y}}^2}{144}=1 \end{gathered}$$

$$\begin{gathered} \mathrm{equation} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{P}(10, 16) \mathrm{is} \\ \frac{{\mathrm{a}}^2\mathrm{x}}{{\mathrm{x}}_1}+\frac{{\mathrm{b}}^2\mathrm{y}}{{\mathrm{y}}_1}={\mathrm{a}}^2+{\mathrm{b}}^2 \\ \Rightarrow \frac{36\mathrm{x}}{10}+\frac{144\mathrm{y}}{16}=180 \\ \Rightarrow 36\left[\frac{\mathrm{x}}{10}+\frac{4\mathrm{y}}{16}\right]=180 \\ \Rightarrow \frac{\mathrm{x}}{10}+\frac{\mathrm{y}}{4}=5 \\ \Rightarrow \frac{2\mathrm{x}+5\mathrm{y}}{20}=5 \\ \Rightarrow 2\mathrm{x}+5\mathrm{y}=100 \end{gathered}$$