A tangent to the curve, $y=f\left(x\right)$ at $P\left(x, y\right)$ meets $x$-axis at $A$ and $y$-axis at $B$. If $AP:BP=1:3$ and $f\left(1\right)=1,$ then the curve also passes through the point

Equation of tangent,

$\begin{array}{l}Y-y=\frac{dy}{dx}(X-x)\\ A\left(x-y\frac{dx}{dy},0\right)\text{}:B\left(0,y-x\frac{dy}{dx}\right)\end{array}$

Given, $AP:BP=1:3$,

Using section formula, we get

$y=\frac{y-x\frac{dy}{dx}}{4}$

$\Rightarrow y-x\frac{dy}{dx}=4y$

$\Rightarrow \frac{dy}{y}+3\frac{dx}{x}=0$

By integrating w.r.t $x$, we get

$\ln \left|y\right|+3\ln \left|x\right|=\ln c$

$\left|y{x}^3\right|=c$

$\because f\left(1\right)=1$

$\therefore c=1$

$\Rightarrow \left|y{x}^3\right|=1$

Hence, the curve passes throught the point $\left(2,\frac{1}{8}\right)$