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anonymous

  • one year ago

Find the vertices and foci of the hyperbola with equation quantity x plus 5 squared divided by 36 minus the quantity of y plus 1 squared divided by 64 equals 1.

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  1. anonymous
    • one year ago
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    @Saisuke<3 Do you know how to do this?

  2. anonymous
    • one year ago
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    @zepdrix @abb0t

  3. zepdrix
    • one year ago
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    \[\Large\rm \frac{(x+5)^2}{6^2}-\frac{(y+1)^2}{8^2}=1\]Let's write it like this so we can easily identify our `a` and `b`. Hmm I don't remember hyperbolas so well :d gonna look it up real quick.

  4. anonymous
    • one year ago
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    yas thats how u set it up!!

  5. anonymous
    • one year ago
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    psh i forgot lol

  6. anonymous
    • one year ago
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    Ok. Thanks for being the one to finally get back to me @zepdrix

  7. anonymous
    • one year ago
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    #20minslater

  8. zepdrix
    • one year ago
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    I'm trying to remember... we open in the direction of the positive one, ya? So since our x is the positive, we open left and right?

  9. anonymous
    • one year ago
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    I believe so

  10. anonymous
    • one year ago
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    It has been a while for me too @zepdrix I can't get through the whole problem by myself

  11. zepdrix
    • one year ago
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    Our "center" is being shifted, so that will affect our vertex point here:|dw:1435533631001:dw|

  12. zepdrix
    • one year ago
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    We would expect the right vertex to be located at \(\Large\rm (6,0)\) but our "center" is shifting that ... `5 to the left` and `1 down`. Do you understand how I came up with those shifts?

  13. anonymous
    • one year ago
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    Yes 5 and -1 from the equation

  14. anonymous
    • one year ago
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    so it would be at 1,-1 for one vertex

  15. zepdrix
    • one year ago
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    MMmmm yah that sounds right. how the other one?

  16. anonymous
    • one year ago
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    Actually that is enough to answer the question but let's continue

  17. anonymous
    • one year ago
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    it WOULD be at -6,0 but it is now at -11,-1

  18. zepdrix
    • one year ago
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    mmm ok good

  19. anonymous
    • one year ago
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    next foci

  20. anonymous
    • one year ago
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    which is what I can't do

  21. zepdrix
    • one year ago
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    ok one sec :) trying to make sense of this picture lol

  22. anonymous
    • one year ago
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    haha

  23. zepdrix
    • one year ago
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    mmm i can't find a good picture to illustrate it... but it seems that the focus points simply satisfy the Pythagorean Theorem.\[\Large\rm c^2=a^2+b^2\]

  24. zepdrix
    • one year ago
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    \[\Large\rm c^2=6^2+8^2\]Ya?

  25. anonymous
    • one year ago
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    ok

  26. zepdrix
    • one year ago
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    Again though, after you find that c value, both the plus and minus, we'll have to apply our shifts

  27. anonymous
    • one year ago
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    so c=10 thats easy

  28. anonymous
    • one year ago
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    so plus and minus as in 10 and -10?

  29. zepdrix
    • one year ago
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    yes, good. one focus point to the left of our `left vertex`, and the other one on the right side of our `right vertex point`.

  30. zepdrix
    • one year ago
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    So normally we would have foci at (-10,0) and (10,0) but we need to apply our shifts

  31. anonymous
    • one year ago
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    yup. so -15,-1 and 5,-1

  32. anonymous
    • one year ago
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    Thanks for the help @zepdrix I will have more later.

  33. zepdrix
    • one year ago
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    cool c:

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