The parabola ${\mathrm{x}}^2=4\mathrm{ay}$ makes an intercept of length $\sqrt{40}$ units on the line $\mathrm{y}=1+2\mathrm{x}$ then a value of 4a is

Given equations 

$$\begin{gathered} {\mathrm{x}}^2=4\mathrm{ay}, \mathrm{y}=1+2x \\ By solving these two equations we get \\ \Rightarrow {\mathrm{x}}^2=4\mathrm{a}(1+2\mathrm{x}) \\ \Rightarrow {\mathrm{x}}^2-8\mathrm{ax}-4\mathrm{a}=0 \end{gathered}$$

$$\begin{gathered} \left|{\mathrm{x}}_1-{\mathrm{x}}_2\right|=\frac{\sqrt{∆}}{\left|\mathrm{a}\right|}=\sqrt{64{\mathrm{a}}^2+16\mathrm{a}}=4\sqrt{4{\mathrm{a}}^2+\mathrm{a}} \\ Length of chord \mathrm{AB}=\left|{\mathrm{x}}_1-{\mathrm{x}}_2\right| \sqrt{1+{\mathrm{m}}^2} \\ \Rightarrow \sqrt{40}=4\sqrt{4{\mathrm{a}}^2+\mathrm{a}} \sqrt{5} \\ \Rightarrow 4\mathrm{a}=-2 \left(\mathrm{or}\right) 1 \end{gathered}$$