If the length of the latus rectum of the ellipse $x^2+4y^2+2x+8y-\lambda =0$ is 4and ‘$\mathcal{l}$’ is the length of major axis, then $\lambda +\mathcal{l}=.....$

Given Ellipse is $x^2+4y^2+2x+8y-\lambda =0$

$\Rightarrow (x+1{)}^2+4(y+1{)}^2=\lambda +5$

$\Rightarrow \frac{(x+1{)}^2}{\lambda +5}+\frac{(y+1{)}^2}{\frac{\lambda +5}{4}}=1$

$\Rightarrow a^2=\lambda +5\text{ and }{b}^2=\frac{\lambda +5}{4}$

$\text{ Given }\frac{2b^2}{a}=4\Rightarrow 2b^2=4a$

$\Rightarrow (\lambda +5{)}^2=64(\lambda +5)\Rightarrow \lambda =59(\because \lambda \ne -5)$

$\text{ Now }{a}^2=64\Rightarrow a=8$

Length of major axis is 2a=16

$\therefore \lambda +l=75$