The eccentricity of the ellipse ${\mathrm{x}}^2+4{\mathrm{y}}^2+2\mathrm{x}+16\mathrm{y}+13=0$ is

$$\begin{gathered} ({\mathrm{x}}^2+2\mathrm{x}+1)+4({\mathrm{y}}^2+4\mathrm{y}+4)-16-1+13=0 \\ {(\mathrm{x}+1)}^2+4{(\mathrm{y}+2)}^2=4 \\ \Rightarrow \frac{{(\mathrm{x}+1)}^2}{4}+\frac{{(\mathrm{y}+2)}^2}{1}=1 \\ where a^2=4,b^2=1(a>b) \\ Now e=\sqrt{\frac{a^2-b^2}{a^2}} \\ =\sqrt{\frac{3}{4} } \\ =\frac{\sqrt{3}}{2} \end{gathered}$$