A tangent to the curve, $$y=f(x)$$ at $$P(x, y)$$ meets $$x-$$axis at $$A$$ and $$y-$$axis at $$B$$. If $$AP:BP=1:3$$ and $$f(1)=1$$, then the curve also passes through the point.
$$\displaystyle\frac{(y-y_1)}{(x-x_1}=f'(x_1)$$
$$y-y_1=f'(x_1)(x-x_1)$$
$$y=0, \displaystyle\frac{-y}{f'(x_1)}=x-x_1$$
$$\Rightarrow x=x_1=\displaystyle\frac{y_1}{f'(x_1)}$$
$$A=\displaystyle \left(x_1-\frac{y_1}{f'(x_1)}, 0\right)$$
$$x=0, y-y_1=f'(x_1)\cdot (-x_1)$$
$$\Rightarrow y=y_1-x_1f'(x_1)$$
$$B=\left(0, y_1-x_1f'(x_1)\right)$$
P divides AB in ratio $$1:3$$
$$\therefore $$
$$x_1=\displaystyle \frac{3\left(x_1-\displaystyle \frac{y_1}{f'(x_1)}\right)}{4}$$
$$y_1=\displaystyle\frac{y_1-x_1f'(x_1)}{4}$$
$$4y_1=y_1x_1f'(x_1)$$
$$f'(x_1)=\displaystyle\frac{-3y_1}{x_1}$$
$$f'(x)=\displaystyle\frac{-3y}{x}$$
$$\displaystyle\frac{dy}{dx}=\frac{-3y}{x}$$
$$\displaystyle\frac{dy}{y}=\frac{-3dx}{x}$$
$$lny=-3ln x+c$$
$$y=k x^{-3}$$
$$y(1)=1$$
$$\Rightarrow k=1$$
$$y=\displaystyle\frac{1}{x^3}$$.
Option $$D$$ satisfies the above function.
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