Length of the shortest normal chord of parabola  $y^2=4ax$  is 

 $$\begin{gathered} A(a{t}_1^2,2a{t}_1)B(a{t}_2^2,2a{t}_2) \\ A{B}^2={\left[a^2(t_1^2-t_2^2)\right]}^2+4a^2{(t_1-t_2)}^2 \\ =a^2{(t_1-t_2)}^2[{(t_1+t_2)}^2+4] \\ =a^2{[t_1+t_1+\frac{2}{t_1}]}^2[\frac{4}{t_1^2}+4] \\ =\frac{16a^2{(1+t_1^2)}^3}{t_1^4} \end{gathered}$$
For maximum (or) minimum
 $f_{\left(t\right)}^1=0$
 $=16a^2\left[\frac{t_1^4\left[3{(1+t_1^2)}^2.2t_1\right]-{(1+t_1^2)}^{3}4t_1^3}{t_1^8}\right]=0$
$t_1=\sqrt{2}$  is point of minima
${\left(AB\right)}_{\min }=\frac{4a}{2}{(1+2)}^{3/2}=2a\sqrt{27}$  units.