anonymous
  • anonymous
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.
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

anonymous
  • anonymous
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.
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
what is the center of this hyperbola?
anonymous
  • anonymous
0,0

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
so (h,k) = (0,0) h = 0 k = 0
anonymous
  • anonymous
hey that makes a cool face 0,0
anonymous
  • anonymous
sorry off topic
jim_thompson5910
  • jim_thompson5910
lol yes it does how far do go from (0,0) to either vertex?
anonymous
  • anonymous
8 units
jim_thompson5910
  • jim_thompson5910
so a = 8 a = half of the length of the transverse axis the transverse axis connects the two vertices and goes through the center
anonymous
  • anonymous
alright
anonymous
  • anonymous
SO what is the formula that we need to plug these into?
jim_thompson5910
  • jim_thompson5910
how is the hyperbola oriented? is it opening left/right? or up/down?
anonymous
  • anonymous
up/down
jim_thompson5910
  • jim_thompson5910
according to this http://www.purplemath.com/modules/hyperbola.htm it says that if the hyperbola opens up/down, then the basic form is \[\Large \frac{(y-k)^2}{a^2} + \frac{(x-h)^2}{b^2} = 1\]
anonymous
  • anonymous
yup
jim_thompson5910
  • jim_thompson5910
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 = ??
anonymous
  • anonymous
16
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\] \[\Large \frac{(y-0)^2}{8^2} - \frac{(x-0)^2}{16^2} = 1\] \[\Large \frac{y^2}{64} - \frac{x^2}{256} = 1\]
jim_thompson5910
  • jim_thompson5910
I had a + in there for some reason. It should have been -
anonymous
  • anonymous
Ok thanks! I have 4 more questions of similar nature if you dont't mind :)
anonymous
  • anonymous
I'll make new questions
jim_thompson5910
  • jim_thompson5910
try the next one out and I'll take a look at the work you have so far on it
anonymous
  • anonymous
Not the same type of question but the same chapter in my book

Looking for something else?

Not the answer you are looking for? Search for more explanations.