If the tangent to the curve y = x³ at the point $\mathrm{p}\left(\mathrm{t}, {\mathrm{t}}^3\right)$ meets. The curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is

$$\begin{gathered} \mathrm{y}={\mathrm{x}}^3\to \left(1\right) \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{x}}^2 \\ \Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{P}}=3{\mathrm{t}}^2 \\ \text{Equation of tangent at }\mathrm{P}\left(\mathrm{t},{\mathrm{t}}^3\right)\text{ is } \end{gathered}$$

$\mathrm{y}-{\mathrm{t}}^3=3{\mathrm{t}}^2(\mathrm{x}-\mathrm{t})\to \left(2\right)$ 
 $$\begin{gathered} \text{Solve (1) }\&\left(2\right) \\ \Rightarrow {\mathrm{x}}^3-{\mathrm{t}}^3=3{\mathrm{t}}^2(\mathrm{x}-\mathrm{t}) \end{gathered}$$

$(\mathrm{x}-\mathrm{t})\left[{\mathrm{x}}^2+\mathrm{xt}+{\mathrm{t}}^2-3{\mathrm{t}}^2\right]=0\Rightarrow (\mathrm{x}-\mathrm{t})\left[{\mathrm{x}}^2+\mathrm{xt}-2{\mathrm{t}}^2\right]=0$

$(\mathrm{x}-\mathrm{t})(\mathrm{x}-\mathrm{t})(\mathrm{x}+2\mathrm{t})=0$ the  abscissae of the point Q is  $\mathrm{x}=-2\mathrm{t}$,

$$\begin{gathered} \mathrm{y}=(-2\mathrm{t}{)}^3=-8{\mathrm{t}}^3 \\ \therefore \mathrm{Q}=\left(-2{\mathrm{t}}_{,}-8{\mathrm{t}}^3\right) \end{gathered}$$

The point which divides $\overline{\mathrm{PQ}}$ in the ratio 1 : 2 = $\left(0,-2{\mathrm{t}}^3\right)$