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

The given parabola is $\mathrm{y}={\mathrm{a}}^3\frac{{\mathrm{x}}^2}{3}+{\mathrm{a}}^2\frac{\mathrm{x}}{2}-2\mathrm{a}$

The above equation can be rewrite as $\frac{{\mathrm{a}}^3}{3}{\left(\mathrm{x}+\frac{3}{4\mathrm{a}}\right)}^2=\left(\mathrm{y}+\frac{35\mathrm{a}}{16}\right)$

The vertex is $\left(-\frac{3}{4a},-\frac{35a}{16}\right)$

Eliminate $a$, to get the locus of vertex of the parabola

suppose that $x=-\frac{3}{4a},y=-\frac{35a}{16}$

Consider the prodcut 

              $\begin{array}{rcl}xy& =& \left(-\frac{3}{4a}\right)\left(-\frac{35a}{16}\right)\\ & =& \frac{105}{64}\end{array}$