Question

Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y equals inverse of plus minus 1 divided by 2x.

Answer 1

@jim_thompson5910

Answer 2

what is the center of this hyperbola?

Answer 3

0,0

Answer 4

yes

Answer 5

so (h,k) = (0,0)
h = 0
k = 0

Answer 6

hey that makes a cool face 0,0

Answer 7

sorry off topic

Answer 8

lol yes it does

how far do go from (0,0) to either vertex?

Answer 9

8 units

Answer 10

so a = 8

a = half of the length of the transverse axis

the transverse axis connects the two vertices and goes through the center

Answer 11

alright

Answer 12

SO what is the formula that we need to plug these into?

Answer 13

how is the hyperbola oriented? is it opening left/right? or up/down?

Answer 14

up/down

Answer 15

according to this

http://www.purplemath.com/modules/hyperbola.htm

it says that if the hyperbola opens up/down, then the basic form is

\[\Large \frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2} = 1\]

Answer 16

yup

Answer 17

on that same page it says that the asymptotes have a slope of +-a/b

focus on say the positive slope


slope = 1/2

a/b = 1/2

8/b = 1/2

b = ??

Answer 18

16

Answer 19

\[\Large \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]

\[\Large \frac{(y-0)^2}{8^2} - \frac{(x-0)^2}{16^2} = 1\]

\[\Large \frac{y^2}{64} - \frac{x^2}{256} = 1\]

Answer 20

I had a + in there for some reason. It should have been -

Answer 21

Ok thanks! I have 4 more questions of similar nature if you dont't mind :)

Answer 22

I'll make new questions

Answer 23

try the next one out and I'll take a look at the work you have so far on it

Answer 24

Not the same type of question but the same chapter in my book