If the length of the latus rectum of the ellipse ${\mathrm{x}}^2+4{\mathrm{y}}^2+2\mathrm{x}+8\mathrm{y}-\mathrm{\lambda }=0$ is 4, and $l$ is the length of its major axis, then $\mathrm{\lambda }+l$ is equal to _______.

$\begin{array}{l}{\mathrm{x}}^2+4{\mathrm{y}}^2+2\mathrm{x}+8\mathrm{y}-\mathrm{\lambda }=0\\ \left({\mathrm{x}}^2+2\mathrm{x}+1\right)+4\left({\mathrm{y}}^2+2\mathrm{y}+1\right)=\mathrm{\lambda }+5\\ \frac{(\mathrm{x}+1{)}^2}{\mathrm{\lambda }+5}+\frac{(\mathrm{y}+1{)}^2}{\frac{\mathrm{\lambda }+5}{4}}=1,(\mathrm{\lambda }\ne -5)\\ \because \mathrm{Length} \mathrm{of} \mathrm{latus} \mathrm{rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}=4\\ \Rightarrow \frac{2(\mathrm{\lambda }+5)}{4}=4\left(\sqrt{\mathrm{\lambda }+5}\right)\Rightarrow \sqrt{\mathrm{\lambda }+5}=8\\ \Rightarrow \mathrm{\lambda }=59\\ \\ l=\mathrm{Length} \mathrm{of} \mathrm{major} \mathrm{axis}=2\mathrm{a}=2\sqrt{\mathrm{\lambda }+5}=2\sqrt{64}=16\\ \Rightarrow \mathrm{\lambda }+\mathrm{\ell }=59+16=75\\ \therefore \mathrm{\lambda }+\mathrm{\ell }=75\end{array}$