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.

- anonymous

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- anonymous

@Saisuke<3 Do you know how to do this?

- anonymous

@zepdrix @abb0t

- 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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## More answers

- anonymous

yas thats how u set it up!!

- anonymous

psh i forgot lol

- anonymous

Ok. Thanks for being the one to finally get back to me @zepdrix

- anonymous

#20minslater

- 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

I believe so

- anonymous

It has been a while for me too @zepdrix I can't get through the whole problem by myself

- zepdrix

Our "center" is being shifted, so that will affect our vertex point here:|dw:1435533631001:dw|

- 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

Yes 5 and -1 from the equation

- anonymous

so it would be at 1,-1 for one vertex

- zepdrix

MMmmm yah that sounds right.
how the other one?

- anonymous

Actually that is enough to answer the question but let's continue

- anonymous

it WOULD be at -6,0 but it is now at -11,-1

- zepdrix

mmm ok good

- anonymous

next foci

- anonymous

which is what I can't do

- zepdrix

ok one sec :) trying to make sense of this picture lol

- anonymous

haha

- 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

\[\Large\rm c^2=6^2+8^2\]Ya?

- anonymous

ok

- zepdrix

Again though, after you find that c value, both the plus and minus,
we'll have to apply our shifts

- anonymous

so c=10
thats easy

- anonymous

so plus and minus as in 10 and -10?

- 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

So normally we would have foci at (-10,0) and (10,0)
but we need to apply our shifts

- anonymous

yup. so -15,-1 and 5,-1

- anonymous

Thanks for the help @zepdrix I will have more later.

- zepdrix

cool c:

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