Normals drawn to y ^ 2 = 4ax at the points where it is intersected...

Let $y = mx + c$, intersect ${y}^{2} = 4ax$ at $$A\left(a{t}_{1}^{2}, 2a{t}_{1}\right)$ and $B\left(a{t}_{2}^{2}, 2a{t}_{2}\right)$.

Then, $\dfrac{2}{{t}_{1} + {t}_{2}} = m$

$⇒ {t}_{1} + {t}_{2} = \dfrac{2}{m}$

Let the foot of another normal be $C\left(a{t}_{3}^{2}, 2a{t}_{3}\right)$

Then, ${t}_{1} + {t}_{2} + {t}_{3} = 0$

${t}_{3} = \left({t}_{1} + {t}_{2}\right) = - \dfrac{2}{m}$

Thus, other foot is \left(\dfrac{4a}{{m}^{2}}, \dfrac{- 4a}{m}\right)$

Hence, option (d) is correct.