## 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.

1. anonymous

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

2. anonymous

@zepdrix @abb0t

3. 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.

4. anonymous

yas thats how u set it up!!

5. anonymous

psh i forgot lol

6. anonymous

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

7. anonymous

#20minslater

8. 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?

9. anonymous

I believe so

10. anonymous

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

11. zepdrix

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

12. 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?

13. anonymous

Yes 5 and -1 from the equation

14. anonymous

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

15. zepdrix

MMmmm yah that sounds right. how the other one?

16. anonymous

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

17. anonymous

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

18. zepdrix

mmm ok good

19. anonymous

next foci

20. anonymous

which is what I can't do

21. zepdrix

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

22. anonymous

haha

23. zepdrix

mmm i can't find a good picture to illustrate it... but it seems that the focus points simply satisfy the Pythagorean Theorem.$\Large\rm c^2=a^2+b^2$

24. zepdrix

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

25. anonymous

ok

26. zepdrix

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

27. anonymous

so c=10 thats easy

28. anonymous

so plus and minus as in 10 and -10?

29. 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.

30. zepdrix

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

31. anonymous

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

32. anonymous

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

33. zepdrix

cool c: