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

As, vertex is at $\left(\pm 6,0\right)$, and it passes through $\left(10,16\right)$,
So, the equation of hyperbola is $\frac{x^2}{36}-\frac{y^2}{144}=1$.
Equation of normal to the hyperbola at $\left(x_1, y_1\right)$ is $\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$

Thus, on putting values, equation of normal is $2x+5y=100$.