Explanation

(d)
Let the slopes be $m,\ m^{2}$
$\therefore m + m^{2} = \frac{- 2h}{b}\ and\ {mm}^{2} = \frac{a}{b}$
$\Rightarrow m^{3} = \left( \frac{a}{b} \right)$
$Now,\ \ m(1 + m) = \frac{- 2h}{b}$
On cubing both sides, we get
$m^{3}\left\lbrack 1 + m^{3} + 3m(1 + m) \right\rbrack = - \frac{{8h}^{3}}{b^{3}}$
$\Rightarrow \frac{a}{b}\left\lbrack 1 + \frac{a}{b} + 3\left( \frac{- 2h}{b} \right) \right\rbrack = \frac{{- 8h}^{3}}{b^{3}}$
$\Rightarrow \frac{b + a}{b} - \frac{6h}{b} = \frac{- 8h^{3}}{{ab}^{2}}$
$\Rightarrow b + a + \frac{{8h}^{3}}{ab} = 6h$
$\Rightarrow \frac{b + a}{h} + \frac{{8h}^{2}}{ab} = 6$