${\mathrm{x}}^2+4{\mathrm{y}}^2+2\mathrm{x}+16\mathrm{y}+\mathrm{k}=0$ represents an  ellipse with ecentricity $\frac{\sqrt{3}}{2}$ for
 

$$\begin{gathered} {\mathrm{x}}^2+2\mathrm{x}+1+4({\mathrm{y}}^2+4\mathrm{y}+4)=-\mathrm{k}+1+16 \\ {(\mathrm{x}+1)}^2+4{(\mathrm{y}+2)}^2=17-\mathrm{k} \\ \frac{{(\mathrm{x}+1)}^2}{17-\mathrm{k}}+\frac{{(\mathrm{y}+2)}^2}{{\displaystyle \frac{17-\mathrm{k}}{4}}}=1 \\ \mathrm{e}=\sqrt{\frac{(17-k)-\left({\displaystyle \frac{17-\mathrm{k}}{4}}\right)}{17-\mathrm{k}}} \\ \Rightarrow \frac{\sqrt{3}}{2}=\sqrt{\frac{(17-k)(1-{\displaystyle \frac{1}{4}})}{17-k}} \\ \therefore \mathrm{for} \mathrm{infinite} \mathrm{values} \mathrm{of} \mathrm{k} \end{gathered}$$