anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Saisuke<3 Do you know how to do this?
anonymous
  • anonymous
@zepdrix @abb0t
zepdrix
  • zepdrix
\[\Large\rm \frac{(x+5)^2}{6^2}-\frac{(y+1)^2}{8^2}=1\]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.

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anonymous
  • anonymous
yas thats how u set it up!!
anonymous
  • anonymous
psh i forgot lol
anonymous
  • anonymous
Ok. Thanks for being the one to finally get back to me @zepdrix
anonymous
  • anonymous
#20minslater
zepdrix
  • zepdrix
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?
anonymous
  • anonymous
I believe so
anonymous
  • anonymous
It has been a while for me too @zepdrix I can't get through the whole problem by myself
zepdrix
  • zepdrix
Our "center" is being shifted, so that will affect our vertex point here:|dw:1435533631001:dw|
zepdrix
  • zepdrix
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?
anonymous
  • anonymous
Yes 5 and -1 from the equation
anonymous
  • anonymous
so it would be at 1,-1 for one vertex
zepdrix
  • zepdrix
MMmmm yah that sounds right. how the other one?
anonymous
  • anonymous
Actually that is enough to answer the question but let's continue
anonymous
  • anonymous
it WOULD be at -6,0 but it is now at -11,-1
zepdrix
  • zepdrix
mmm ok good
anonymous
  • anonymous
next foci
anonymous
  • anonymous
which is what I can't do
zepdrix
  • zepdrix
ok one sec :) trying to make sense of this picture lol
anonymous
  • anonymous
haha
zepdrix
  • zepdrix
mmm i can't find a good picture to illustrate it... but it seems that the focus points simply satisfy the Pythagorean Theorem.\[\Large\rm c^2=a^2+b^2\]
zepdrix
  • zepdrix
\[\Large\rm c^2=6^2+8^2\]Ya?
anonymous
  • anonymous
ok
zepdrix
  • zepdrix
Again though, after you find that c value, both the plus and minus, we'll have to apply our shifts
anonymous
  • anonymous
so c=10 thats easy
anonymous
  • anonymous
so plus and minus as in 10 and -10?
zepdrix
  • zepdrix
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`.
zepdrix
  • zepdrix
So normally we would have foci at (-10,0) and (10,0) but we need to apply our shifts
anonymous
  • anonymous
yup. so -15,-1 and 5,-1
anonymous
  • anonymous
Thanks for the help @zepdrix I will have more later.
zepdrix
  • zepdrix
cool c:

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