Given, $x^{n} y^{m}=a^{m+n}, \quad m, n>0$
Taking logarithm on both sides, we get
$$
\begin{gathered}
\log \left(x^{n} y^{m}\right)=\log a^{m+n} \\
\Rightarrow \quad \log x^{n}+\log y^{m}=(m+n) \log a \\
\Rightarrow n \log x+m \log y=(m+n) \log a
\end{gathered}
$$
On differentiating Eq. (i) w.r.t. ' $x$ ', we get
$\frac{\mathrm{n}}{\mathrm{x}}+\frac{\mathrm{m}}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=0$
$\Rightarrow \quad \frac{\mathrm{m}}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{n}}{\mathrm{x}}$
$\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=-\left(\frac{\mathrm{n}}{\mathrm{m}}\right)\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$
$\therefore$ Length of subtangent
$=\frac{y}{d y / d x}$
$=\frac{y}{-\left(\frac{n}{m}\right)\left(\frac{y}{x}\right)}$
$=\frac{-m x}{n}$
$\therefore$ Length of subtangent at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\frac{\mathrm{m}}{\mathrm{n}}\left|\mathrm{x}_{1}\right|$