The distance between the foci of the ellipse $\mathrm{x}=3\mathrm{Cos} \mathrm{\theta }; \mathrm{y}=4\mathrm{Sin} \mathrm{\theta }$ is

$$\begin{gathered} Given equations are \\ \frac{\mathrm{x}}{3}=\cos \mathrm{\theta } \mathrm{and} \frac{\mathrm{y}}{4}=\sin \mathrm{\theta } \\ Now {\left(\frac{\mathrm{x}}{3}\right)}^2+{\left(\frac{\mathrm{y}}{4}\right)}^2={\cos }^2\theta +{\sin }^2\theta \\ \Rightarrow \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{16}=1 \\ \mathrm{distance} \mathrm{between} \mathrm{foci}==2\mathrm{be} \\ =2\sqrt{16-9} \\ =2\sqrt{7} \end{gathered}$$