Prove that any point on the ellipse whose foci are `(-1,0)` and `(7,0)` and eccentricity is `1/2` is `(3+8costheta,4sqrt(3)sintheta),theta in Rdot`

Explanation

The foci are (-1,0 ) and (7,0) .
The distance between the foci is 2ae=8 or ae=4
Since e=1/2, we have a=8
Now, `b^(2)=a^(2)(1-e^(2))`
or `b^(2)=48`
or `b=4sqrt(3)`.
The center of the ellipse is the midopoint of the joining two foci. Therefore, the coordinates of the center are (3,0)
Hence its equation is
`((x-3)^(2))/(8^(2))+((y-0)^(2))/((4sqrt(3)))=1 " "(1)`
Thus, hte parameter coordinates of a point on (1) are `(3+8 cos theta, 4 sqrt(3)sin theta)`