Explanation

(c)
Given curve is $x^{3} - 8a^{2}\ y = 0.$
On differentiating w.r.t. $x$, we get
$3x^{2} - 8a^{2}\ \frac{dy}{dx} = 0$
$\Rightarrow \ \ \frac{dy}{dx} = \frac{3x^{2}}{8a^{2}}$
$\therefore$ Slope of the normal $= - \frac{1}{\left( \frac{dy}{dx} \right)} = - \frac{1}{\frac{3x^{2}}{8a^{2}}} = - \frac{8a^{2}}{3x^{2}}$
Given, $\frac{- 8a^{2}}{3x^{2}} = \frac{- 2}{3}\ \ \Rightarrow \ \ x^{2} = 4a^{2}\ \Rightarrow \ \ \ x = \pm 2a$
At $x = \pm 2a,\ \ y = \pm a$
$\therefore\ (x,\ y) = (2a,\ a)$