Given the cubic parabola $y=a x^{3}$, the tangent line at point $A$ intersects the $Y$-axis at point $B$ and the curve itself at another point $C$. Determine the geometric locus of the midpoints of segments $A B$ and $B C$ as the position of point $A$ changes!
$1^{0}$. Let the abscissa of point $A$ be $p$; its ordinate $a p^{3}$. The slope of the tangent line drawn at point $A$ is given by the value of $y'=\frac{dy}{dx}$ at $x=p$. Since $y'=3 a x^{2}$, its value at point $A$ is $3 a p^{2}$. The equation of the tangent line at point $A$ is:

$$
y-a p^{3}=3 a p^{2}(x-p) \quad \text { or } \quad 3 a p^{2} x-y-2 a p^{3}=0
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_1ccca3a874c40427f4c8g-1.jpg?height=488&width=442&top_left_y=350&top_left_x=836)

For the intersection of this tangent line with the $Y$-axis, $x=0$. Thus, the ordinate of $B$ is: $-2 a p^{3}$. The coordinates of the midpoint $F_{1}$ of the distance between $A(p, a p^{3})$ and $B(0, -2 a p^{3})$ are:

$$
x_{1}=\frac{p}{2}, \quad y_{1}=\frac{a p^{3}-2 a p^{3}}{2}=-\frac{a p^{3}}{2}
$$

Eliminating $p$ from these two equations (by substituting $p=2 x$), and dropping the index, we obtain the equation of the curve traced by the midpoint of the distance $AB$:

$$
y=-4 a x^{3} \square
$$

$2^{0}$. Now let's find the coordinates of the intersection point of line $E$ with the curve. We need to solve the system of equations:

$$
y=a x^{3} \quad \text { and } \quad 3 a p^{2} x-y-2 a p^{3}=0
$$

Eliminating $y$ from these two equations, we get:

$$
\begin{aligned}
a x^{3}-3 a p^{2} x+2 a p^{3} & =0 \\
\text { or } \quad x^{3}-3 p^{2} x+2 p^{3} & =0
\end{aligned}
$$

This equation has $p$ as a double root, meaning the left side must be divisible by $(x-p)^{2}$. Indeed,

$$
\left(x^{3}-3 p^{2} x+2 p^{3}\right)=(x-p)^{2}(x+2 p)
$$

Thus, for the intersection point $C$ of the tangent line $E$ and the curve: $x_{3}+2 p=0, x_{3}=-2 p$; its ordinate is: $y_{3}=a(-2 p)^{3}=-8 a p^{3}$. For the midpoint $F_{2}$ of the distance between $B(0, -2 a p^{3})$ and $C(-2 p, -8 a p^{3})$:

$$
x'=-p, \quad y'=\frac{-2 a p^{3}-8 a p^{3}}{2}=-5 a p^{3}
$$

Eliminating $p$ from these two equations, the relationship between the coordinates of the midpoint of the distance $BC$ is:

$$
y=5 a x^{3}{ }^{2}
$$

Fehér György (Fazekas Mihály g. VIII. o. Debrecen).[^0]


[^0]:    ${ }^{1} \mathrm{If}$ the symmetric image of the curve $y=a x^{3}$ with respect to the $X$-axis is constructed and then the ordinate of every point is quadrupled, the curve $y=-4 a x^{3}$ is obtained.

    ${ }^{2} \mathrm{If} \mathrm{the}$ ordinate of every point on the curve $y=a x^{3}$ is quintupled, the curve $y=5 a x^{3}$ is obtained.