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

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jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Ooh... 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437111744452:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that what it would look like?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and no, I don't have the formulas for those

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Basically, the rectangle would just be touching the hyperbola, and the ellipse would fit perfectly inside, but I know drawing here is difficult. :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Let me find those equations for the foci.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Focus of hyperbola: \[a^2 + b^2 = f^2\] Focus of ellipse: \[a^2  b^2 = f^2\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Let'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?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Great! We can do the same thing for the ellipse. What equation should we start with?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a^2 + b^2 = c_e^2\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Good! Just make that + into a  sign (it's a different equation for the ellipse) :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, that's right i forgot

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1And I made a mistake by saying, we can do the "same thing", not exactly the same I guess. :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1In any case, you will get \[c_e = \sqrt{a^2  b^2}\] Yay, we are getting closer!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, great, so how do I started finding the distance between the two?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Good question. That seems a little tough at first, but let's see if we draw the situation.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437112353871:dw

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Well, now I can see that the distance between them is easy to find!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm still confused on how to do that

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Well, for one thing, let's only talk about the positive focuses (foci). So just look at the right hand side.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1You can see that c_h is sitting far to the right, while c_e is sitting closer to the middle.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Just to get you thinking... maybe c_h = 5 and c_e = 3. What would be the distance between them if that was true?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Exactly! 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, okay. So it would be \[\sqrt{a^2+b^2}\sqrt{a^2b^2}\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1That 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
 one year ago
Best ResponseYou've already chosen the best response.0oh, that actually wasn't too difficult! Thank you so much

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1That's always a great feeling when something crazy turns out to be easier than you thought. Your welcome! Good luck!
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