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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.

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  1. anonymous
    • one year ago
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  2. jtvatsim
    • one year ago
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    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?

  3. anonymous
    • one year ago
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    |dw:1437111744452:dw|

  4. anonymous
    • one year ago
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    is that what it would look like?

  5. anonymous
    • one year ago
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    and no, I don't have the formulas for those

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

  7. jtvatsim
    • one year ago
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    Let me find those equations for the foci.

  8. anonymous
    • one year ago
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    ok

  9. jtvatsim
    • one year ago
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    Focus of hyperbola: \[a^2 + b^2 = f^2\] Focus of ellipse: \[a^2 - b^2 = f^2\]

  10. jtvatsim
    • one year ago
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    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?

  11. anonymous
    • one year ago
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    ok

  12. jtvatsim
    • one year ago
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    Great! We can do the same thing for the ellipse. What equation should we start with?

  13. anonymous
    • one year ago
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    \[a^2 + b^2 = c_e^2\]

  14. jtvatsim
    • one year ago
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    Good! Just make that + into a - sign (it's a different equation for the ellipse) :)

  15. anonymous
    • one year ago
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    oh, that's right i forgot

  16. jtvatsim
    • one year ago
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    And I made a mistake by saying, we can do the "same thing", not exactly the same I guess. :)

  17. jtvatsim
    • one year ago
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    In any case, you will get \[c_e = \sqrt{a^2 - b^2}\] Yay, we are getting closer!

  18. anonymous
    • one year ago
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    ok, great, so how do I started finding the distance between the two?

  19. jtvatsim
    • one year ago
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    Good question. That seems a little tough at first, but let's see if we draw the situation.

  20. anonymous
    • one year ago
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    ok

  21. jtvatsim
    • one year ago
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    |dw:1437112353871:dw|

  22. jtvatsim
    • one year ago
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    Well, now I can see that the distance between them is easy to find!

  23. anonymous
    • one year ago
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    i'm still confused on how to do that

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

  25. anonymous
    • one year ago
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    ok

  26. jtvatsim
    • one year ago
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    You can see that c_h is sitting far to the right, while c_e is sitting closer to the middle.

  27. anonymous
    • one year ago
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    yes

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

  29. anonymous
    • one year ago
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    2?

  30. jtvatsim
    • one year ago
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    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.

  31. anonymous
    • one year ago
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    Oh, okay. So it would be \[\sqrt{a^2+b^2}-\sqrt{a^2-b^2}\]

  32. jtvatsim
    • one year ago
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    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.

  33. anonymous
    • one year ago
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    oh, that actually wasn't too difficult! Thank you so much

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

  35. anonymous
    • one year ago
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    Yeah, thanks!

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