The tangent to the curve x³-ay²=0 at an arbitrary point T

The tangent to the curve $x^{3}-ay^{2}=0$ at an arbitrary point $T$ meets the curve again at $S$. Show that the $x$-axis divides the segment $\overline{ST}$ in the ratio $8:1$.

Step 1: Parametrise the curve

Use $x=at^{2}$, $y=at^{3}$, since then $x^{3}=a^{3}t^{6}=a(at^{3})^{2}=ay^{2}$. Let $T=(at^{2},at^{3})$.

Step 2: Slope of the tangent

Differentiating $x^{3}=ay^{2}$ implicitly, $3x^{2}=2ay\,y' \Rightarrow y'=\dfrac{3x^{2}}{2ay}$; at $T$, \[k = \frac{3(at^{2})^{2}}{2a\cdot at^{3}} = \frac{3t}{2}. \tag{1}\]

Step 3: Tangent at $T$

\[ y-at^{3} = \frac{3t}{2}(x-at^{2}) \;\Longrightarrow\; y = \frac{3t}{2}x - \frac{at^{3}}{2}. \tag{2} \]

Step 4: Second intersection $S$

Substituting (2) into $x^{3}=ay^{2}$ and multiplying by $4$: \[4x^{3} - 9at^{2}x^{2} + 6a^{2}t^{4}x - a^{3}t^{6} = 0. \tag{3}\]

Since $x=at^{2}$ is a double root (the point of tangency), this factors as

\[ 4x^{3}-9at^{2}x^{2}+6a^{2}t^{4}x-a^{3}t^{6} = (x-at^{2})^{2}(4x-at^{2}) = 0, \tag{4} \] so $x_{S}=at^{2}/4$; substituting into the tangent (2) gives $y_{S}=-at^{3}/8$, i.e.

\[ S = \left(\frac{at^{2}}{4},\; -\frac{at^{3}}{8}\right). \tag{5} \]

Step 5: Where $Ox$ meets $\overline{ST}$

Parametrise the segment by $y(\lambda)=(1-\lambda)y_{S}+\lambda\,y_{T}$ and set $y=0$: \[0 = -\frac{at^{3}}{8}(1-\lambda)+at^{3}\lambda \;\Longrightarrow\; -\frac{1-\lambda}{8}+\lambda=0 \;\Longrightarrow\; \lambda = \frac{1}{9}. \tag{6}\]

So the crossing point $Q$ satisfies $\overline{SQ}:\overline{QT}=\lambda:(1-\lambda)=1:8$, equivalently $\overline{TQ}:\overline{QS}=8:1$.

Step 6: Verification

\[Q\text{: } x_Q=\tfrac89\cdot\tfrac{at^2}{4}+\tfrac19\cdot at^2=\tfrac{at^2}{3},\ y_Q=0\ \checkmark\]

\[|QT|_y:|SQ|_y = at^3 : \tfrac{at^3}{8} = 8:1\ \checkmark\]