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Two promotional stands are to be placed in an ellipse-shaped room–one at each foci–and are 3 units from the center. If the room is 8 units wide, what is its length?

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Final answer:

The length of the ellipse-shaped room is determined by the formula for an ellipse, c² = a² - b². Given the width (2a) as 8 units and the distance to each focus (c) as 3 units, we solve for b and find that the room's length is 2√7 units, approximately 5.29 units.

Step-by-step explanation:

To find the length of the ellipse-shaped room, we can use the properties of an ellipse. The width of the room, which corresponds to the major axis (2a), is given as 8 units. The distance from the center to the focal points (f) is 3 units for each promotional stand. In an ellipse, the relationship between the semi-major axis (a), the semi-minor axis (b), and the focal length (f) is c² = a² - b² where c is the distance from the center to a focus.

Since the width of the room is the length of the major axis, we have 2a = 8 units, so a = 4 units. We are given that c = 3 units. Plugging these into the ellipse formula we get 3² = 4² - b², which simplifies to b² = 4² - 3² = 16 - 9 = 7. Therefore, b = √7 units.

The length of the room is twice the semi-minor axis, which is 2b, so the length is 2√7 units. For simplification, since √7 is approximately 2.645, the length is approximately 2 * 2.645 = 5.29 units.

User Bryan B
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