SOLUTION: A whispering gallery has a semielliptical ceiling that is 12m high and 40m long how high is the ceiling above the two foci?

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Question 1088120: A whispering gallery has a semielliptical ceiling that is 12m high and 40m long how high is the ceiling above the two foci?
Answer by ikleyn(52812) About Me  (Show Source):
You can put this solution on YOUR website!
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From the condition, the major semi-axis of the ellipse has the length of a = 20 m.

The length of the minor semi-axis is b = 12 m.

The linear eccentricity is c = sqrt%28a%5E2+-+b%5E2%29 = sqrt%2820%5E2-12%5E2%29 = sqrt%28400+-+144%29 = sqrt%28256%29 = 16.


So the foci are located on the major axis (which is the horizontal line at the floor) 

at the distance of 16 m from the center point.


Taking the center point as the origin of the coordinate system,
the standard equation of the ellipse is 

x%5E2%2F400 + y%5E2%2F144 = 1.              (1)


Then y = +/- 12%2Asqrt%281-x%5E2%2F400%29.     (2)


To calculate the elevation at the focus, we must substitute x = 16 into the formula (2) and calculate y:

y = 12%2Asqrt%281-16%5E2%2F400%29 = 12%2Asqrt%28144%2F400%29 = %2812%2A12%29%2F20 = 36%2F5 = 7.2 m.


Answer.  The ceiling above the two foci is at 7.2 m.

Solved.

The prerequisite for this solution is the lesson
    - Ellipse definition, canonical equation, characteristic points and elements
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic
"Conic sections: Ellipses. Definition, major elements and properties. Solved problems".