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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.

Álvaro Lozano-Robledo
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Eric Brown
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3 Answers3

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Consider this sketch

enter image description here

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 $\dfrac{x^2}{75^2}+\dfrac{y^2}{45^2}=1$

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

Andrei
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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.

Tojra
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