The tangent to the curve ay = x³ at a point A meets

The tangent to the curve $ay = x^{3}$ at a point $A$ meets the curve again at a point $B$. Find the locus of the midpoints of the segment $\overline{AB}$.

Step 1: Parametrise $A$ and write the tangent

Let $A=(t,\,t^{3}/a)$ be an arbitrary point on $ay=x^{3}$. Since $y'=3x^{2}/a$, the slope at $A$ is $k=3t^{2}/a$, and the tangent is

\[ y-\frac{t^{3}}{a} = \frac{3t^{2}}{a}(x-t) \;\Longrightarrow\; y = \frac{3t^{2}}{a}x - \frac{2t^{3}}{a}. \tag{1} \]

Step 2: The second intersection $B$

Substituting $y=x^{3}/a$ into (1) and multiplying by $a$: \[x^{3} - 3t^{2}x + 2t^{3} = 0. \tag{2}\]

Since $x=t$ is the point of tangency, it is a double root; factoring, \[x^{3}-3t^{2}x+2t^{3} = (x-t)^{2}(x+2t) = 0, \tag{3}\] so the third root is $x=-2t$, giving

\[ B = \left(-2t,\; -\frac{8t^{3}}{a}\right). \tag{4} \]

Step 3: Midpoint $M$ of $\overline{AB}$

\[ M = \left(\frac{t+(-2t)}{2},\; \frac{\dfrac{t^{3}}{a}+\left(-\dfrac{8t^{3}}{a}\right)}{2}\right) = \left(-\frac{t}{2},\; -\frac{7t^{3}}{2a}\right). \tag{5} \]

Step 4: Eliminate the parameter $t$

Let $M=(X,Y)$. From (5), $X=-t/2 \Rightarrow t=-2X$; substituting into $Y$, \[Y = -\frac{7t^{3}}{2a} = -\frac{7(-2X)^{3}}{2a} = \frac{56X^{3}}{2a} = \frac{28X^{3}}{a}, \qquad \text{i.e.}\qquad aY = 28X^{3}. \tag{6}\]

Step 5: Verification

\[A,B \text{ on the curve: } a\cdot t^3/a=t^3\ \checkmark;\quad a\cdot(-8t^3/a)=-8t^3=(-2t)^3\ \checkmark\]

\[\text{Double root at } x=t\text{: } (x-t)^2(x+2t)\big|_{x=t}=0\ \checkmark\]

\[\text{Midpoint on locus: } a\!\left(-\tfrac{7t^3}{2a}\right)=-\tfrac{7t^3}{2}=28\!\left(-\tfrac{t}{2}\right)^3\ \checkmark\]