Explanation

(d)
Given curves are $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{12} = 1$
$\Rightarrow \ \ \ \ \ \frac{2x}{a^{2}} + \frac{2y}{12} \bullet \frac{dy}{dx} = 0$
$\Rightarrow \ \ \ \ \frac{dy}{dx} = - \frac{12x}{a^{2}y} = m_{1}$ (say)
And $y^{3} = 8x \Rightarrow 3y^{2}\frac{dy}{dx} = 8$
$\Rightarrow \ \ \ \ \frac{dy}{dx} = \frac{8}{3y^{2}} = m_{2}\$ (say)
For $\theta = \frac{\pi}{2},1 + m_{1}m_{2} = 0$
$\Rightarrow \ \ 1 + \left( \frac{- 12x}{a^{2}y} \right)\left( \frac{8}{3y^{2}} \right) = 0$
$\Rightarrow \ \ \ 3a^{2}(8x) - 96x = 0$
$\Rightarrow \ \ \ \ a^{2} = 4$