The parametric representation of a point on the ellipse whose foci are \((-1,0)\) and \((7,0)\) and eccentricity \(1 / 2\), is

Distance between two foci, \(2 a e=7+1=8\)
\(\therefore\) \(
a e=4
\)
\(
\Rightarrow
\)
\(a=8\)
\(
\left(\because e=\frac{1}{2} \text { given }\right)
\)
Now,
\(
\begin{array}{l}
b^{2}=a^{2}\left(1-e^{2}\right)=64\left(1-\frac{1}{4}\right) \\
b^{2}=48 \Rightarrow b=4 \sqrt{3}
\end{array}
\)
\(\therefore\)
Since, the centre of the ellipse is the mid point of the line joining two foci, therefore the coordinates of the centre are \((3,0)\). \(\therefore\) Its equation is
\(
\frac{(x-3)^{2}}{8^{2}}+\frac{(y-0)^{2}}{(4 \sqrt{3})^{2}}=1
\)
Hence, the parametric coordinates of a point on Eq. (i) are \((3+8 \cos \theta, 4 \sqrt{3} \sin \theta)\).