The point of intersection of the latus rectum and axis of the parabola $y^2+4x+2y-8=0$ is the focus of the parabola.

$y^2+4x+2y-8=0$

$\Rightarrow y^2+2y+1=-4x+8+1$

$\Rightarrow {\left(y+1\right)}^2=4\left(-1\right)\left(x-\frac{9}{4}\right)$

We know that the focus of the parabola ${\left(y-k\right)}^2=4a\left(x-h\right)$ is $S\equiv \left(a+h,k\right)$.

So, the focus of the given parabola is $S\equiv \left(-1+\frac{9}{4},-1\right)\equiv \left(\frac{5}{4},-1\right)$.