If the length of the latus rectum of the ellipse x 2 + 4 y 2 + 2 x + 8 y - λ = 0 is 4 , and l is the…

Given, 

$x^2+4y^2+2x+8y-\lambda =0$ is a ellipse

Now on rearranging the terms we get,

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

We know that length of latus rectum is $\frac{2b^2}{a}=4$ 

So, $\frac{2\left(\lambda +5\right)}{4\left({\displaystyle \sqrt{\lambda +5}}\right)}=4$

$\Rightarrow \frac{2\left(\lambda +5\right)}{4}=4\left(\sqrt{\lambda +5}\right)$

$\Rightarrow \left(\lambda +5\right)=8\left(\sqrt{\lambda +5}\right)$

$\Rightarrow {\left(\lambda +5\right)}^2=64\left(\lambda +5\right)$

$\Rightarrow {\lambda }^2+25+10\lambda =64\left(\lambda +5\right)$

$\Rightarrow {\lambda }^2-54\lambda -295=0$

$\Rightarrow \lambda =59 \text{or }-5$

But $\lambda \ne -5$

So, $l=2a=2\sqrt{\lambda +5}=2\sqrt{64}=16$

$\Rightarrow \lambda +l=59+16=75$