If the tangent at any point $P$ on the curve $x^m{y}^n=a^{m+n}(mn\ne 0)$ meets the coordinate axes in $A$ and $B$ then show that $AP : BP$ is a constant.

Given curve is $x^m{y}^n=a^{m+n}$

Taking log on b.s.

$\log \left(x^m{y}^n\right)=\log \left(a^{m+n}\right)$

$\log x^m+\log y^n=(m+n)\log a$

$m\log x+n\log y=(m+n)\log a$

Differentiating w.r.to $x$

$m\cdot \frac{1}{x}+n\cdot \frac{1}{y}\frac{dy}{dx}=0$

$\Rightarrow \frac{n}{y}\frac{dy}{dx}=-\frac{m}{x}\Rightarrow \frac{dy}{dx}=-\frac{my}{nx}$

Slope of tangent at $P\left(x_1,y_1\right)$

$m={\left(\frac{dy}{dx}\right)}_{P\left(x_1,y_1\right)}=\frac{-m{y}_1}{n{x}_1}$

Equation of tangent at ${ }^{\text{'}}{p}^{\text{'} }\left(x_1,y_1\right)$ is

$\boxed{y-y_1=m\left(x-x_1\right)}$

$\Rightarrow y-y_1=\frac{-m{y}_1}{n{x}_1}\left(x-x_1\right)$

$\Rightarrow \frac{n}{y_1}\left(y-y_1\right)=\frac{-m}{x_1}\left(x-x_1\right)$

$\Rightarrow \frac{ny}{y_1}-n=\frac{-mx}{x_1}+m\Rightarrow \frac{mx}{x_1}+\frac{ny}{y_1}=m+n$

$\Rightarrow \frac{mx}{(m+n)x_1}+\frac{ny}{(m+n)y_1}=1$

$\Rightarrow \frac{x}{\left(\frac{m+n}{m}\right)x_1}+\frac{y}{\left(\frac{m+n}{n}\right)y_1}=1$

The tangent cuts the $x$-axis at

$A\left(\left(\frac{m+n}{m}\right)x_1,0\right)$ and $\mathrm{y}\text{-axis at }B\left(0,\left(\frac{m+n}{n}\right)y_1\right)$

‘$P$’ divides $AB$ in ratio $AP : PB=x_1-x:x-x_2$

$=x_1\left(\frac{m+n}{m}\right)-x_1:x_1-0$

$=x_1\left[\frac{m+n}{m}-1\right]:x_1=\frac{m+n-m}{m}:1$

$AP:PB=\frac{n}{m}:1\Rightarrow n:m$