If a hyperbola passes through the point

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

If a hyperbola passes through vertices at $(\pm 6,0)$, then

$\therefore \quad a=6$

As hyperbola passes through the point $P(10,16)$

$\therefore \quad \frac{100}{36}-\frac{256}{b^{2}}=1 \Rightarrow b^{2}=144$

$\therefore \quad$ Required hyperbola is $\frac{x^{2}}{36}-\frac{y^{2}}{144}=1$

$\therefore \quad$ Required hyperbola is $\frac{x^{2}}{36}-\frac{y^{2}}{144}=1$

Equation of normal is $\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}=a^{2}+b^{2}$

$\therefore \quad$ At $\mathrm{P}(10,16)$ normal is

$\frac{36 x}{10}+\frac{144 y}{16}=36+144$

$\therefore \quad 2 x+5 y=100 .$