Explanation

(d)
Given, $y + 2a = \frac{a^{2}}{3}\left( x^{2} + \frac{3}{2a}x \right)$
$\Rightarrow \ y + 2a = \frac{a^{3}}{3}\left( x + \frac{3}{4a} \right)^{2} - \frac{9}{16^{2}} \times \frac{a^{3}}{3}$
$\Rightarrow \left( y + \frac{35a}{16} \right) = \frac{a^{3}}{3}\left( x + \frac{3}{4a} \right)^{2}$
Thus, the vertices of parabola is $\left( - \frac{3}{4a}, - \frac{35a}{16} \right)$
Let $h = - \frac{3}{4a}$ and $k = - \frac{35a}{16}$
$\therefore hk = \frac{105}{64}$
Hence, the locus of vertices of a parabola is $xy = \frac{105}{64}$