The locus of the vertices of the family of parabolas $$y = \dfrac{a^3 x^2}{3} + \dfrac{a^2 x}{2} - 2a$$ is

The given family of parabolas is
$$y=\frac{a^{3}x^{2}}{3}+\frac{a^{2}x}{2}-2a \qquad (a\in\mathbb{R})$$

For any parabola written as $$y=Ax^{2}+Bx+C$$ the vertex $$V(x_{v},y_{v})$$ is obtained from

$$x_{v}=-\frac{B}{2A}, \qquad y_{v}=C-\frac{B^{2}}{4A}$$

Here
$$A=\frac{a^{3}}{3},\qquad B=\frac{a^{2}}{2},\qquad C=-2a$$

Step 1: x-coordinate of the vertex
$$x=-\frac{B}{2A}=-\frac{\dfrac{a^{2}}{2}}{2\left(\dfrac{a^{3}}{3}\right)}=-\frac{3}{4a}$$

Step 2: y-coordinate of the vertex
$$$ \begin{aligned} y&=C-\frac{B^{2}}{4A}\\ &=-2a-\frac{\left(\dfrac{a^{2}}{2}\right)^{2}}{4\left(\dfrac{a^{3}}{3}\right)}\\ &=-2a-\frac{a^{4}/4}{4a^{3}/3}\\ &=-2a-\frac{3a}{16}\\ &=-\frac{35a}{16} \end{aligned} $$$

Thus every vertex has coordinates
$$\bigl(x,\;y\bigr)=\left(-\frac{3}{4a},\;-\frac{35a}{16}\right).$$

Step 3: Eliminate the parameter $$a$$

From $$x=-\dfrac{3}{4a}\; \Rightarrow\; a=-\dfrac{3}{4x}$$

Substitute this value in the expression for $$y$$:
$$y=-\frac{35}{16}\left(-\frac{3}{4x}\right)=\frac{105}{64}\,\frac{1}{x}$$

Re-arranging, we obtain the locus

$$xy=\frac{105}{64}.$$

Hence the locus of the vertices is $$xy=\dfrac{105}{64}$$.

Option A which is: $$xy = \dfrac{105}{64}$$