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

$\begin{array}{l}\text{ Vertex is at }(\pm 6,0)\\ \therefore a=6\end{array}$

$\text{ Let the hyperbola is }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Putting point P(10, 16) on the hyperbola

$\begin{array}{l}\frac{100}{36}-\frac{256}{b^2}=1 \Rightarrow b^2=144\\ \therefore \text{ Hyperbola is }\frac{x^2}{36}-\frac{y^2}{144}=1\end{array}$

$\text{ Equation of normal is }\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$

$\therefore \text{ Putting }\left(x_1,y_1\right)=(10,16),\text{ we get }2x+5y=100$