Explanation

(c)
Let any point on the parabola be $(at^{2},\ 2at)$
If the equation of parabola is $y^{2} = 4ax$, then focus is $(a,0)$
Let the focus of a point be $(\alpha,\ \beta)$ if it is a mid point
$\therefore\alpha = \frac{at^{2} + a}{2},\ \beta = \frac{2at + 0}{2}$
$\Rightarrow 2\alpha = at^{2} + a,\ \ \beta = at$
$\therefore 2\alpha = a\left( \frac{\beta}{a} \right)^{2\ } + a$
$\Rightarrow 2a\alpha = \beta^{2} + a^{2}$
$\Rightarrow \beta^{2} = - a^{2} + 2a\alpha$
$\Rightarrow \beta^{2} = \frac{4a}{2}\left( \alpha - \frac{a}{2} \right)$
$\therefore$ The locus is $y^{2} = \frac{4a}{2}\left( x - \frac{a}{2} \right)$
The directrix is $X = - \frac{a}{2}$
$\Rightarrow x - \frac{a}{2} = - \frac{a}{2}$
$\Rightarrow x = 0$