The locus of the vertices of the family of the parabola’s $y=\frac{a^3{x}^2}{3}+\frac{a^{2}x}{2}-2ais$

$$\begin{gathered} \text{ Given family of parabolas is }y=\frac{a^3{x}^2}{3}+\frac{a^{2}x}{2}-2a \\ \Rightarrow \frac{y}{a^3/3}=x^2+\frac{a^2}{2}\cdot \frac{3}{a^3}x-\frac{2a}{a^3/3} \\ \begin{array}{l}\Rightarrow \frac{3y}{a^3}+\frac{6a}{a^3}=x^2+2\left(\frac{3}{4a}\right)x\Rightarrow {\left(x+\frac{3}{4a}\right)}^2 \\ =\frac{3y}{a^3}+\frac{6}{a^2}+\frac{9}{16a^2}\Rightarrow {\left(x+\frac{3}{4a}\right)}^2=\frac{3y}{a^3}+\frac{105}{16a^2}\\ \Rightarrow {\left(x+\frac{3}{4a}\right)}^2=\frac{3}{a^3}\left(y+\frac{35a}{16}\right)\Rightarrow \text{ Vertex }=\left(\frac{-3}{4a},\frac{-35a}{16}\right) \\ \text{ Let }\left(x_1,y_1\right)\text{ be a point in the locus.} \\ \text{Then }{x}_1=(-3/4a),y_1=(-35a/16) \\ \text{ Now }{x}_1{y}_1=\left(-\frac{3}{4a}\right)\left(-\frac{35a}{16}\right)\Rightarrow x_1{y}_1=\frac{105}{64} \\ \therefore \text{ Equation to the locus is }xy=\frac{105}{64}\end{array} \end{gathered}$$