Graph a hyperbola with foci at (0, ±10) and vertices at (0, ±8) using graphing technology.

- anonymous

Graph a hyperbola with foci at (0, ±10) and vertices at (0, ±8) using graphing technology.

- schrodinger

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- hba

@brookester6 can help you on this :P

- anonymous

umm yeah JAS!

- anonymous

wow a lot of people on one question lol party

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## More answers

- anonymous

well, I did tag a lot of people. I really need help.

- anonymous

:( everyone gone.

- Loser66

I witnessed many people stopped by to help you, why did they go all?

- anonymous

they, left, It's too hard of a question

- anonymous

@brookester6 said just a second, and then left.

- Loser66

not that girl!! I think because you called too many people. so, this one thinks that the other's duty. hahahaa

- anonymous

:( aw man, i just need an equation

- anonymous

Let's start from scratch again

- Loser66

ok, my question is " what does graphing technology?" I always graph by hand, don't know how to use "technology"

- anonymous

lol using a website, but I just need an equation, I already have a website to use

- Loser66

so, you can solve it by yourself? no need to stay ? right? actually, I am a super lazy guy. If you don't need help anymore, I glad to hear that. hehehe

- anonymous

no, i need to have the numbers put into an equation, I don't know how to do that

- Loser66

shoot.what is your equation? for hyperbola?

- anonymous

foci at (0, ±10) and vertices at (0, ±8) Make this into an equation

- Loser66

remember that foci (0,+-c) and vertices (0,+-a) and c^2 = a^2 + b^2( use this form to find out b) and put everything into the standard form of hyperbola
\[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\]

- Loser66

and draw the asymptotes by
\[y= (+-)\frac{a}{b}*x\]

- Loser66

that's it

- anonymous

10^2=8^2+b^2

- Loser66

remember that if your foci is (0,+- c) your hyperbola has the shape like|dw:1369325453179:dw|

- Loser66

so b^2 = 100 - 64 = 36
so ?

- anonymous

100=64+b^2
36=b^2
6=b

- Loser66

\[\frac{y^2}{64}-\frac{x^2}{36}=1\]
your equation form is a^2 and b^2 , you need a, b when you have to figure out the asymptotes only. to get the form of hyperbola, no need to get that far

- Loser66

got what i mean?

- anonymous

ok, but how do i find x or y?

- Loser66

why you have to know x, y?

- anonymous

oh wait, that's it

- Loser66

yeap!!!

- Loser66

if you have to sketch the graph, you have to sketch the asymptotes, too. but it's quite easy when you have a, b, just draw the lines out.

- anonymous

I got the graph

- anonymous

http://www3.wolframalpha.com/Calculate/MSP/MSP52421f29bdchhib18c60000436031h3b4eaa8g7?MSPStoreType=image/gif&s=37&w=200.&h=204.&cdf=RangeControl

- anonymous

But, I have online class

- anonymous

girl?

- anonymous

no you say girl at the end of sentence, I CONFUSE

- Loser66

i think you are a girl

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