Answer
This is an ellipse, with the main axis on the x-axis:
the center is $(-2,0)$
the foci are $(-4,0)$ and $(0,0)$
the vertices are $(-2\pm\sqrt{5},0)$.
Work Step by Step
We group like terms
$(x^{2}+4x)+5y^{2}=1$
We complete the square
$(x^{2}+4x+4)+5y^{2}=1+4$
$(x+2)^{2}+5y^{2}=5$
Divide with 5
$\displaystyle \frac{(x+2)^{2}}{5}+y^{2}=1$
This has the form $\quad \quad \displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad (a\gt b)$
Foci on the x-axis$, a=\sqrt{5}, \quad b=1$
Center-to-focus distance: $\quad c=\sqrt{a^{2}-b^{2}}=\sqrt{5-1}=2$
Foci: $\quad(\pm 2, 0)$
Vertices: $\quad (\pm\sqrt{5}, 0)$
When $x$ is replaced with $(x+2)$, a shift to the left is made.
$(x,y)\rightarrow(x-2,y)$
(No vertical shift $\Rightarrow$ the main axis remains on the x-axis.)
The center shifts to $(0-2,0)=(-2,0)$
The foci shift to $(\pm 2-2, 0)\Rightarrow (-4,0)$ and $(0,0)$
The vertices shift to $(-2\pm\sqrt{5}, 0)$
