Since length of the chord intercepted from the line
$y=m x +c$ by the parabola $x^{2}=4 a y$ is
$4 \sqrt{a\left(1+m^{2}\right)\left(c +a m^{2}\right)}$
So, $\sqrt{40}=4 \sqrt{a(1+4)(1+a(4))}$
$40 =16(5 a(1+4 a))$
$\Rightarrow 1 =2 a(1+4 a)$
$\Rightarrow 8 a^{2}+2 a-1=0$
$\Rightarrow 8 a^{2}+4 a-2 a-1=0$
$\Rightarrow 4 a(2 a+1)-1(2 a+1)=0$
$\Rightarrow a=\frac{1}{4}$ or $-\frac{1}{2}$
$\Rightarrow 4 a=1$ or$-2$