Question

A "whispering room" is one with an elliptically-arched ceiling. If someone stands at one focus of the ellipse and whispers something to his friend, the dispersed sound waves are reflected by the ceiling and concentrated at the other focus, allowing people across the room to clearly hear what he said. Suppose such gallery has a ceiling reaching twenty feet above the five-foot-high vertical walls at its tallest point (so the cross-section is half an ellipse topping two vertical lines at either end), and suppose the foci of the ellipse are thirty feet apart. What is the height of the ceiling above each "whispering point"?

• A. The ceiling is $20$ feet above the floor.
• B. The ceiling is $21$ feet above the floor.
• C. The ceiling is $23$ feet above the floor.
• D. None of the above

Answer 1

Answer: (B)

The ceiling is $21$ feet above the floor.

Let us consider the ellipse over the origin that is $(h.k)=(0,5)$

The foci are thirty feet apart, so they are 15 units to the either side of the centre.

IN particular $c=15$. SInce the elliptical part of the room's cross section is twenty feet high above the center, and since this "shorter" direction is the semi-major axis, then $b=20.$The equation $b^2=a^2-c^2$ gives

$400=a^2-225$, so $a^2=625$, $a=25$

Therefore the equation of the elliptical ceiling is

$\frac{(x-0{)}^2}{625}+\frac{(y-5{)}^2}{400}=1$

We should find the height of the ceiling above the foci. Let us take the focus to the right of the centre. The height will be the $y$- value of the ellipse when $x=15$.

$\frac{(15{)}^2}{625}+\frac{(y-5{)}^2}{400}=1$

$\frac{9}{25}+\frac{(y-5{)}^2}{400}=1$

$144+(y-5{)}^2=400$

$(y-5{)}^2=256$

$y-5=\pm 16$

$y=21$

The ceiling is 21 feet above the floor.