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

Let the point of tangency be $$P(x,y)$$ on the unknown curve $$y=f(x)$$. Denote the slope of the tangent at this point by $$m=\dfrac{dy}{dx}\;.$$ The equation of the tangent in point-slope form is

$$y-y=m(x-x)\;.\qquad -(1)$$

(In every occurrence, $$x$$ and $$y$$ without subscripts represent the coordinates of $$P$$.)

The x-intercept $$A$$ is obtained by putting $$y=0$$ in $$-(1)$$:

$$0-y=m(x_A-x)\;$$

$$\Longrightarrow \;x_A=x-\dfrac{y}{m}\;.\qquad -(2)$$

Thus $$A\bigl(x-\dfrac{y}{m},\,0\bigr)\;.$$

The y-intercept $$B$$ is obtained by putting $$x=0$$ in $$-(1)$$:

$$y_B-y=m(0-x)=-mx\;$$

$$\Longrightarrow \;y_B=y-mx\;.\qquad -(3)$$

Thus $$B\bigl(0,\,y-mx\bigr)\;.$$

We now need the (positive) lengths of the tangent segments $$AP$$ and $$BP$$. Writing the vectors from $$P$$ to these intercepts,

$$\overrightarrow{PA}=\Bigl(-\dfrac{y}{m},\,-y\Bigr),\qquad \overrightarrow{PB}=(-x,\,-mx)\;.$$

Hence

$$|AP|=\sqrt{\left(-\dfrac{y}{m}\right)^2+(-y)^2} =\sqrt{\dfrac{y^2}{m^2}+y^2} =|y|\sqrt{\dfrac{1}{m^2}+1} =\dfrac{|y|}{|m|}\sqrt{1+m^2}\;,\qquad -(4)$$

$$|BP|=\sqrt{(-x)^2+(-mx)^2} =\sqrt{x^2+m^2x^2} =|x|\sqrt{1+m^2}\;.\qquad -(5)$$

The ratio of these positive lengths is therefore

$$\dfrac{|AP|}{|BP|} =\dfrac{\dfrac{|y|}{|m|}\sqrt{1+m^2}}{|x|\sqrt{1+m^2}} =\dfrac{|y|}{|m||x|}\;.\qquad -(6)$$

The statement of the problem gives $$AP:BP=1:3$$, i.e. $$|AP|/|BP|=1/3\;.$$ Substituting this in $$-(6)$$ we get

$$\dfrac{|y|}{|m||x|}=\dfrac13\;.$$

Because $$x>0,\;y>0$$ at $$P(1,1)$$, we can drop the absolute values of $$x$$ and $$y$$, but the sign of $$m$$ (the slope) is not known a-priori. Hence

$$\left|\dfrac{y}{mx}\right|=\dfrac13 \;\Longrightarrow\;m=\pm\dfrac{3y}{x}\;.\qquad -(7)$$

Equation $$-(7)$$ links the slope $$m$$ with the coordinates $$x,y$$ on the curve. Replacing $$m$$ by the derivative $$\dfrac{dy}{dx}$$ gives the differential equations

$$\dfrac{dy}{dx}=\dfrac{3y}{x}\quad\text{or}\quad \dfrac{dy}{dx}=-\dfrac{3y}{x}\;.\qquad -(8)$$

Both are separable. Treating the positive sign first:

$$\dfrac{dy}{y}=3\,\dfrac{dx}{x}\;,$$

$$\int\dfrac{dy}{y}=\int 3\,\dfrac{dx}{x}\;,$$

$$\ln y=3\ln x+C_1\;,$$

$$y=e^{C_1}x^3\;.$$

Writing $$k=e^{C_1}$$ (a non-zero constant), we have

$$y=kx^3\;.\qquad -(9)$$

For the negative sign in $$-(8)$$ we obtain

$$\dfrac{dy}{y}=-3\,\dfrac{dx}{x}\;,$$

$$\ln y=-3\ln x+C_2\;,$$

$$y=e^{C_2}x^{-3}\;,$$

$$y=\dfrac{k}{x^3}\quad(k=e^{C_2})\;.\qquad -(10)$$

Thus two one-parameter families of curves satisfy the required tangent property:

$$y=kx^3\quad\text{or}\quad y=\dfrac{k}{x^3}\;.\qquad -(11)$$

The curve is known to pass through the given point $$P(1,1)$$. Substituting $$x=1,\;y=1$$ in $$-(11)$$ gives

$$1=k(1)^3\;\Longrightarrow\;k=1\;.$$

This value of $$k$$ works in both families, so the two possible explicit curves are

$$y=x^3\quad\text{and}\quad y=\dfrac1{x^3}\;.\qquad -(12)$$

We must now determine which of the four listed points actually lies on one (and hence on any) of these curves.

Testing the point $$\bigl(\dfrac{1}{3},24\bigr):$$ If $$y=x^3$$ then $$y=\bigl(\dfrac13\bigr)^3=\dfrac1{27}\neq24$$. If $$y=1/x^3$$ then $$y=27\neq24$$. This point is not on either curve.

Testing the point $$\bigl(\dfrac12,4\bigr):$$ For $$y=x^3$$ we get $$y=\dfrac1{8}\neq4$$, and for $$y=1/x^3$$ we get $$y=8\neq4$$. This point is not on either curve.

Testing the point $$(2,\dfrac18):$$ For $$y=x^3$$ we get $$y=8\neq\dfrac18$$, but for $$y=1/x^3$$ we obtain $$y=\dfrac1{8}$$, which matches perfectly. Hence this point does lie on the second curve.

Testing the point $$(3,\dfrac1{28}):$$ For $$y=x^3$$ we get $$y=27\neq\dfrac1{28}$$; for $$y=1/x^3$$ we get $$y=\dfrac1{27}\neq\dfrac1{28}$$. So this point is also rejected.

Only the point $$(2,\dfrac18)$$ is satisfied by one of the admissible curves, namely $$y=\dfrac1{x^3}\;.$$

Hence, the correct answer is Option C.