At the beginning of this lesson, I'd mentioned that ellipses have real-world applications. Naturally, these applications can be turned into word problems. You'll often be dealing with half-ellipses, forming some sort of dish or arc; the word problems will refer to a bridge support, or an arched ceiling, or something similar.
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The important thing to remember with ellipses is that sounds or lights directed at one focus will bounce around and reflect back out to the other focus. The foci will generally be the important parts of an ellipse that is applied in a practical situation.
Since the ceiling is half of an ellipse (the top half, specifically), and since the foci will be on a line between the tops of the "straight" parts of the side walls, the foci will be five feet above the floor, which sounds about right for people talking and listening: five feet high is close to face-high on most adults.
I'll center my ellipse above the origin, so the center is at (h, k) = (0, 5). The foci are thirty feet apart, so they're 15 units to either side of the center; this means that c = 15.
Since the elliptical part of the room's cross-section is twenty feet above the center, and since this "shorter" direction is the semi-minor axis, then b = 20.
The equation b2 = a2 − c2 gives me 400 = a2 − 225, so a2 = 625. Then the equation for the elliptical ceiling is:
I need to find the height of the ceiling above the foci. I prefer positive numbers, so I'll look at the focus to the right of the center. The height (from the ellipse's central line through its foci, up to the ceiling) will be the y-value of the ellipse when x = 15:
144 + (y − 5)2 = 400
(y − 5)2 = 256
y − 5 = ±16
y = 21
(Since I'm looking for the height above, not the depth below, I ignored the negative solution to the quadratic equation.)
The ceiling is 21 feet above the floor.
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The lowest altitude will be at the vertex closer to the Earth; the highest altitude will be at the other vertex. Since I need to measure these altitudes from the focus, I need to find the value of c.
b2 = a2 − c2
c2 = a2 − b2 = 44202 − 44162 = 35,344
Then c = 188. If I set the center of my ellipse at the origin and make this a wider-than-tall ellipse, then I can put the Earth's center at the point (188, 0).
(This means, by the way, that there isn't much difference between the circumference of the Earth and the path of the satellite. The center of the elliptical orbit is actually inside the Earth, and the ellipse, having an eccentricity of e = 188/4420, or about 0.04, is pretty close to being a circle.)
The vertex closer to the end of the ellipse containing the Earth's center will be at 4420 units from the ellipse's center, or 4420 − 188 = 4232 units from the center of the Earth. Since the Earth's radius is 3960 units, then the altitude is 4232 − 3960 = 272.
The other vertex is 4420 + 188 = 4608 units from the Earth's center, giving me an altitude of 4608 − 3960 = 648 units.
minimum altitude: 272 miles
maximum altitude: 648 miles