Answer:
Let the coordinate of focus be [tex](\pm c , 0)[/tex]
As per the statement: The base of an auditorium is in the form of an eclipse 200 feet long and 100 feet wide.
⇒Length of Major axis=base of an auditorium = 200 feet and Length of a minor axis=wide of a auditorium = 100 ft
Semi-major axis (a) = 100 ft and
semi-minor axis(b) = 50 ft
Then, by an equation:
[tex]c^2 = a^2-b^2[/tex]
Solve for c:
Substitute the given values we have;
[tex]c^2=(100)^2-(50)^2[/tex]
Simplify:
[tex]c^2 = 7500[/tex]
or
[tex]c=\sqrt{7500} = 86.6025404[/tex] ft
Distance between the foci is, [tex]2c = 2 \cdot 86.6025404 = 173.205081[/tex]
Therefore, the distance between the foci to the nearest 10th of a foot is, 173.2 ft
