Tangent is drawn at any point P of a curve which passes through (1,1)…

The equation of the tangent at $P(x, y)$ is $Y-y=\frac{d y}{d x}(X-x)$

It cuts the coordinate axes at

$A\left(x-y \frac{d x}{d y}, 0\right)$ and $\left(0, y-x \frac{d y}{d x}\right)$

It is given that $A P: B P=3: 1$

∴   $\frac{3 \times 0+x-y \frac{d x}{d y}}{3+1}=x$ and, $\frac{3\left(y-x \frac{d y}{d x}\right)+1 \times 0}{3+1}=y$

$\Rightarrow x-y \frac{d x}{d y}=4 x$ and $3 y-3 x \frac{d y}{d x}=4 y$

$\Rightarrow \frac{d y}{d x}=-\frac{y}{3 x} \Rightarrow 3 x \frac{d y}{d x}+y=0$

$\Rightarrow \frac{3}{y} d y+\frac{1}{x} d x=0$

Integrating, we get

$y^3 x=C$              .......(i)

It passes through $(1,1)$

∴  $C=1$

Putting $C=1$ in (i), we obtain $y^3 x=1$ as the equation of the curve. Clearly, it passes through $(1 / 8,2)$.