The equation of the chord, having mid-point as $(x_1, y_1)$, of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is given by
$T=S_{1}\,...\left(i\right)$
where, $T=\frac{xx_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}-1 $ and
$S_{1}=\frac{x^{2}_{1}}{a^{2}}-\frac{y^{2}_{1}}{b^{2}}-1$
According to the question,
$\left(x_{1}, y_{1}\right) = \left(5, 3\right)$ and $a^{2} = 16, b^{2} = 25$
as $25x^{2} - 16y^{2} = 400$
$\Rightarrow \frac{x^{2}}{16}-\frac{y^{2}}{25}=1\,\therefore \frac{5x}{16}-\frac{3y}{25}=\frac{25}{16}-\frac{9}{25}\left[Using \left(i\right)\right]$
$\Rightarrow 125x - 48y = 625 - 144 \Rightarrow 125x - 48y = 481$