1. Write the ellipse in standard form

Given: $x^{2}+4y^{2}+2x+8y-\lambda=0$

Group and complete squares: $x^2+2x+4(y^2+2y)-\lambda=0$

$x^2+2x=(x+1)^2-1$ $y^2+2y=(y+1)^2-1$

So, $(x+1)^2-1+4\big[(y+1)^2-1\big]-\lambda=0$

$(x+1)^2+4(y+1)^2-(\lambda+5)=0$

Hence, $(x+1)^2+4(y+1)^2=\lambda+5$

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

Since $\lambda+5>0$ for an ellipse, the larger denominator is $a^2=\lambda+5, \qquad b^2=\frac{\lambda+5}{4}$ So the major axis is along the $x$-direction.

2. Use the latus rectum formula

For an ellipse, $\text{length of latus rectum}=\frac{2b^2}{a}$

Given it is $4$, so $\frac{2b^2}{a}=4$

Now, $b^2=\frac{\lambda+5}{4}, \qquad a=\sqrt{\lambda+5}$

Thus, $\frac{2\cdot \frac{\lambda+5}{4}}{\sqrt{\lambda+5}}=4$

$\frac{\lambda+5}{2\sqrt{\lambda+5}}=4$

$\frac{\sqrt{\lambda+5}}{2}=4$

$\sqrt{\lambda+5}=8$

$\lambda+5=64$

$\lambda=59$

3. Find the length of the major axis

Major axis length is $l=2a=2\sqrt{\lambda+5}=2\cdot 8=16$

4. Compute $\lambda+l$

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

Final Answer

$\boxed{75}$