Official solution
Given parabola is \[y = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2a\]
\[ \Rightarrow y = {{{a^3}} \over 3}\left( {{x^3} + {3 \over {2a}} + x + {9 \over {16{a^2}}}} \right) - {{3a} \over {16}} - 2a\]
\[ \Rightarrow y + {{35a} \over {16}} = {{{a^3}} \over 3}{\left( {x + {3 \over {4a}}} \right)^2}\]
\[\therefore\] Vertex of parabola is \[\left( {{{ - 3} \over {4a}},{{ - 35a} \over {16}}} \right)\]
To find locus of this vertex,
\[x = {{ - 3} \over {4a}}\,\,\] and \[\,\,y = {{ - 35a} \over {16}}\]
\[ \Rightarrow a = {{ - 3} \over {4x}}\,\,\] and \[a = - {{16y} \over {35}}\]
\[ \Rightarrow {{ - 3} \over {4x}} = {{ - 16y} \over {35}} \Rightarrow 64xy = 105\]
\[ \Rightarrow xy = {{105} \over {64}}\] which is the required locus.