Explanation

(d)
Given that equation of parabola is $y^{2} = 8x$
$\Rightarrow a = 2$
We know, if the normal at point $(at_{1}^{2},\ 2at_{1})$ is passing through the point on the parabola $(at_{2}^{2},\ 2at_{2})$, then $t_{2} = - t_{1} - \frac{2}{t_{1}}$
Given point is $(2,\ 4)$
$\Rightarrow at_{1}^{2} = 2$
$\Rightarrow t_{1} = 1$
$\therefore t_{2} = - 1 - \frac{2}{1} = - 3$
The other end will be $(at_{2}^{2},\ 2at_{2})ie$, $(18,\ - 12)$