Find the equation and height of an elliptical whispering room

Question

The room is 150 feet long and the distance from the center of the room to the foci is 60 feet.

Finding $a^2$ is easy its $$2a=150$$ $$a=75$$ $$a^2=5625$$

but where I get lost is finding $b^2$, I know I shouldn't look at the answer before solving but I was stuck for 10 mins. trying to figure it out.

Anyway $b=45$ and I have no idea how to calucate that because what am I suppose to use the $60$ feet from the foci for?

I thought it was $$2b=60$$ $$b=30$$ $$b^2=900$$ but that doesn't work, then I thought it should be $$2b=120$$ $$b=60$$ $$b^2=3600$$ and again I get stuck.

Answer 1

Consider this sketch

We know

• $BF_1+BF_2= AF_1+AF_2 = 2AO = 150$ as this is an ellipse with a fixed combined distance from the foci to a point on the edge

• and that $OF_1=OF_2=60$

• so $OA^2 =75^2$ as you found

• while $OB^2 = 75^2-60^2 = 45^2$ by Pythagoras

so the height (or width?) is $2OB=90$

and the equation of the room might be $\dfrac{x^2}{75^2}+\dfrac{y^2}{45^2}=1$

Answer 2

You can use $c^2=a^2-b^2$, so $b=\sqrt{a^2-c^2}=\sqrt{75^2-60^2}=15\sqrt{5^2-4^2}=15\cdot3=45$

Answer 3

It is simple, just you need to know more about ellipse. In ellipse, the distance of centre from focus is $ae$, where $e=\sqrt{1-(\frac b a)^2}$. Now it is easy.

e is called eccentricity of the ellipse.