Gven, equation of the parabola
$x^{2}=a y$
ie $y=\frac{x^{2}}{a}$
and the equation of line
$y-2 x=1$
$y=2 x+1$
$\frac{x^{2}}{a}=2 x+1$
$x^{2}=2 a x+a$
$x^{2}-2 a x-a=0$
$\begin{aligned} x &=\frac{2 a \pm \sqrt{4 a^{2}+4 a}}{2} \\ &=\frac{2\left(a \pm \sqrt{a^{2}+a}\right)}{2}=a \pm \sqrt{a^{2}+a} \end{aligned}$
Points of intersection of the parabola and the line are
$$
\left(a+\sqrt{a^{2}+a}, \left(2\left(a+\sqrt{a^{2}+a}\right)+1\right)\right)
$$
and $\quad\left(a-\sqrt{a^{2}+a}, \left(2 ( a-\sqrt{a^{2}+a})+1\right)\right)$
$\because$ Distance between the points $=\sqrt{40}$ $\therefore \quad \sqrt{40}=\sqrt{\left(2 \sqrt{a^{2}+a}\right)^{2}+\left(4 \sqrt{a^{2}+a}\right)^{2}}$
$\Rightarrow \quad 40=4\left(a^{2}+a\right)+16\left(a^{2}+a\right)$
$\Rightarrow \quad 2=a^{2}+a \Rightarrow a=1-2$