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

Question

hba · Answer

@brookester6 can help you on this :P

anonymous · Answer

umm yeah JAS!

anonymous · Answer

wow a lot of people on one question lol party

anonymous · Answer

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

anonymous · Answer

:( everyone gone.

loser66 · Answer

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

anonymous · Answer

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

anonymous · Answer

@brookester6 said just a second, and then left.

loser66 · Answer

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

anonymous · Answer

:( aw man, i just need an equation

anonymous · Answer

Let's start from scratch again

loser66 · Answer

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

anonymous · Answer

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

loser66 · Answer

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 · Answer

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

loser66 · Answer

shoot.what is your equation? for hyperbola?

anonymous · Answer

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

loser66 · Answer

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 · Answer

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

loser66 · Answer

that's it

anonymous · Answer

10^2=8^2+b^2

loser66 · Answer

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