The tangent to y = x³ at a point P meets the x-axis at S
The tangent to $y = x^{3}$ at a point $P$ meets the $x$-axis at $S$; the orthogonal projection of $P$ on the $x$-axis is $T$. Show that $\overline{OS} = 2\,\overline{ST}$, where $O$ is the origin.
Explanation of Method
The subtangent of the cubic is a third of the abscissa.
Step 1: The tangent
At $P = \bigl(x_{0},x_{0}^{3}\bigr)$, $x_{0}\neq0$, the slope is $y' = 3x_{0}^{2}$, so
\[ y-x_{0}^{3} = 3x_{0}^{2}\bigl(x-x_{0}\bigr) \;\Longrightarrow\; y = 3x_{0}^{2}x-2x_{0}^{3}. \tag{1} \]
Step 2: The two points on the axis
Setting $y = 0$ in (1): \[3x_{0}^{2}x = 2x_{0}^{3} \;\Longrightarrow\; x = \frac{2x_{0}}{3}, \qquad S = \left(\frac{2x_{0}}{3},\,0\right),\] while the projection of $P$ is $T = \bigl(x_{0},0\bigr)$.
Step 3: The two lengths
All three points lie on the $x$-axis, so their signed distances are their abscissas: \[\overline{OS} = \frac{2x_{0}}{3}, \qquad \overline{ST} = x_{0}-\frac{2x_{0}}{3} = \frac{x_{0}}{3}.\]
Step 4: Verification
\[x_{0} = 3:\ \text{tangent } y = 27x-54,\ S = \bigl(2,0\bigr),\ T = \bigl(3,0\bigr):\ \overline{OS} = 2,\ \overline{ST} = 1 \;\;\checkmark\]
\[x_{0} = -1:\ S = \left(-\tfrac23,0\right),\ T = \bigl(-1,0\bigr):\ \overline{OS} = -\tfrac23,\ \overline{ST} = -\tfrac13 \;\;\checkmark\]
\[\text{The subtangent } \overline{ST} = \frac{y_{0}}{y'} = \frac{x_{0}^{3}}{3x_{0}^{2}} = \frac{x_{0}}{3} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\frac{\overline{OS}}{\overline{ST}} = \frac{2x_{0}/3}{x_{0}/3} = 2 \;\Longrightarrow\; \overline{OS} = 2\,\overline{ST} \quad\text{for every } x_{0}\neq0\,}\]
Remark
Remark. The last line is the general fact: for $y = x^{n}$ the subtangent is $\dfrac{x_{0}}{n}$, so $S$ divides $OT$ in the ratio $\bigl(n-1\bigr):1$ and $\overline{OS} = \dfrac{n-1}{n}\,\overline{OT}$. For $n = 3$ that is $\overline{OS} = \tfrac23\overline{OT} = 2\overline{ST}$; for the parabola $n = 2$ it gives the familiar bisection $\overline{OS} = \overline{ST}$.