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

Given family of parabolas is  $y=\frac{a^3{x}^2}{3}+\frac{a^{2}x}{2}-2a\Rightarrow \frac{y}{a^3/3}=x^2+\frac{a^2}{2}\cdot \frac{3}{a^3}x-\frac{2a}{a^3/3}$

$\Rightarrow \frac{3y}{a^3}+\frac{6a}{a^3}=x^2+2\left(\frac{3}{4a}\right)x\Rightarrow {\left(x+\frac{3}{4a}\right)}^2=\frac{3y}{a^3}+\frac{6}{a^2}+\frac{9}{16a^2}\Rightarrow {\left(x+\frac{3}{4a}\right)}^2=\frac{3y}{a^3}+\frac{105}{16a^2}$

$\Rightarrow {\left(x+\frac{3}{4a}\right)}^2=\frac{3}{a^3}\left(y+\frac{35a}{16}\right)\Rightarrow$ Vertex $=\left(\frac{-3}{4a},\frac{-35a}{16}\right)$

Let $\left(x_1,y_1\right)$ be a point in the locus. Then $x_1=(-3/4a),y_1=(-35a/16)$

Now $x_1{y}_1=\left(-\frac{3}{4a}\right)\left(-\frac{35a}{16}\right)\Rightarrow x_1{y}_1=\frac{105}{64}$

$\therefore$ Equation to the locus is $xy=\frac{105}{64}$