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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.
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Saisuke<3 Do you know how to do this?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large\rm \frac{(x+5)^2}{6^2}\frac{(y+1)^2}{8^2}=1\]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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yas thats how u set it up!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok. Thanks for being the one to finally get back to me @zepdrix

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3I'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
 one year ago
Best ResponseYou've already chosen the best response.0It has been a while for me too @zepdrix I can't get through the whole problem by myself

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Our "center" is being shifted, so that will affect our vertex point here:dw:1435533631001:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3We 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes 5 and 1 from the equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it would be at 1,1 for one vertex

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3MMmmm yah that sounds right. how the other one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually that is enough to answer the question but let's continue

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it WOULD be at 6,0 but it is now at 11,1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which is what I can't do

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3ok one sec :) trying to make sense of this picture lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3mmm 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
 one year ago
Best ResponseYou've already chosen the best response.3\[\Large\rm c^2=6^2+8^2\]Ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Again though, after you find that c value, both the plus and minus, we'll have to apply our shifts

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so plus and minus as in 10 and 10?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3yes, 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
 one year ago
Best ResponseYou've already chosen the best response.3So normally we would have foci at (10,0) and (10,0) but we need to apply our shifts

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yup. so 15,1 and 5,1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the help @zepdrix I will have more later.
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