anonymous
  • anonymous
I need help with this question about hyperbolas and ellipses. The problem in in the attached file.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
jtvatsim
  • 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
  • anonymous
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anonymous
  • anonymous
is that what it would look like?
anonymous
  • anonymous
and no, I don't have the formulas for those
jtvatsim
  • 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
  • jtvatsim
Let me find those equations for the foci.
anonymous
  • anonymous
ok
jtvatsim
  • jtvatsim
Focus of hyperbola: \[a^2 + b^2 = f^2\] Focus of ellipse: \[a^2 - b^2 = f^2\]
jtvatsim
  • 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
  • anonymous
ok
jtvatsim
  • jtvatsim
Great! We can do the same thing for the ellipse. What equation should we start with?
anonymous
  • anonymous
\[a^2 + b^2 = c_e^2\]
jtvatsim
  • jtvatsim
Good! Just make that + into a - sign (it's a different equation for the ellipse) :)
anonymous
  • anonymous
oh, that's right i forgot
jtvatsim
  • jtvatsim
And I made a mistake by saying, we can do the "same thing", not exactly the same I guess. :)
jtvatsim
  • jtvatsim
In any case, you will get \[c_e = \sqrt{a^2 - b^2}\] Yay, we are getting closer!
anonymous
  • anonymous
ok, great, so how do I started finding the distance between the two?
jtvatsim
  • jtvatsim
Good question. That seems a little tough at first, but let's see if we draw the situation.
anonymous
  • anonymous
ok
jtvatsim
  • jtvatsim
|dw:1437112353871:dw|
jtvatsim
  • jtvatsim
Well, now I can see that the distance between them is easy to find!
anonymous
  • anonymous
i'm still confused on how to do that
jtvatsim
  • jtvatsim
Well, for one thing, let's only talk about the positive focuses (foci). So just look at the right hand side.
anonymous
  • anonymous
ok
jtvatsim
  • jtvatsim
You can see that c_h is sitting far to the right, while c_e is sitting closer to the middle.
anonymous
  • anonymous
yes
jtvatsim
  • 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
  • anonymous
2?
jtvatsim
  • 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
  • anonymous
Oh, okay. So it would be \[\sqrt{a^2+b^2}-\sqrt{a^2-b^2}\]
jtvatsim
  • 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
  • anonymous
oh, that actually wasn't too difficult! Thank you so much
jtvatsim
  • jtvatsim
That's always a great feeling when something crazy turns out to be easier than you thought. Your welcome! Good luck!
anonymous
  • anonymous
Yeah, thanks!

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