- anonymous

I need help with this question about hyperbolas and ellipses. The problem in in the attached file.

- schrodinger

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- anonymous

##### 1 Attachment

- jtvatsim

Ooh... this just looks like too much fun! :)
You should probably start by drawing a picture of the situation for yourself.
Also,
Do you happen to have the formulas for the foci of hyperbolas and ellipses?

- anonymous

|dw:1437111744452:dw|

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## More answers

- anonymous

is that what it would look like?

- anonymous

and no, I don't have the formulas for those

- jtvatsim

Basically, the rectangle would just be touching the hyperbola, and the ellipse would fit perfectly inside, but I know drawing here is difficult. :)

- jtvatsim

Let me find those equations for the foci.

- anonymous

ok

- jtvatsim

Focus of hyperbola:
\[a^2 + b^2 = f^2\]
Focus of ellipse:
\[a^2 - b^2 = f^2\]

- jtvatsim

Let's start with the focus of the hyperbola since that's what your first bullet point is. Using the questions terminology we have \[a^2 + b^2 = c_h^2\] which is just \[c_h = \sqrt{a^2 + b^2}\]
OK so far?

- anonymous

ok

- jtvatsim

Great! We can do the same thing for the ellipse. What equation should we start with?

- anonymous

\[a^2 + b^2 = c_e^2\]

- jtvatsim

Good! Just make that + into a - sign (it's a different equation for the ellipse) :)

- anonymous

oh, that's right i forgot

- jtvatsim

And I made a mistake by saying, we can do the "same thing", not exactly the same I guess. :)

- jtvatsim

In any case, you will get \[c_e = \sqrt{a^2 - b^2}\] Yay, we are getting closer!

- anonymous

ok, great, so how do I started finding the distance between the two?

- jtvatsim

Good question. That seems a little tough at first, but let's see if we draw the situation.

- anonymous

ok

- jtvatsim

|dw:1437112353871:dw|

- jtvatsim

Well, now I can see that the distance between them is easy to find!

- anonymous

i'm still confused on how to do that

- jtvatsim

Well, for one thing, let's only talk about the positive focuses (foci). So just look at the right hand side.

- anonymous

ok

- jtvatsim

You can see that c_h is sitting far to the right, while c_e is sitting closer to the middle.

- anonymous

yes

- jtvatsim

Just to get you thinking... maybe c_h = 5 and c_e = 3. What would be the distance between them if that was true?

- anonymous

2?

- jtvatsim

Exactly! I would need to walk 2 units to get from one to the other. Now you found that by doing 5 - 3, right?
You could have done c_h - c_e then to get the distance.
But we have formulas for c_h and c_e! So all we do is subtract them and we get the distance.

- anonymous

Oh, okay. So it would be \[\sqrt{a^2+b^2}-\sqrt{a^2-b^2}\]

- jtvatsim

That right! And it makes sense too. a^2 + b^2 will give a bigger square root than a^2 - b^2 so our subtraction won't be negative or something weird.

- anonymous

oh, that actually wasn't too difficult! Thank you so much

- jtvatsim

That's always a great feeling when something crazy turns out to be easier than you thought. Your welcome! Good luck!

- anonymous

Yeah, thanks!

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