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Mathematics 19 Online
OpenStudy (anonymous):

Find an equation in standard form for the hyperbola with

OpenStudy (anonymous):

OpenStudy (mertsj):

Well, I see the center is at the origin, do you see that?

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

Is a=3 and b=4?

OpenStudy (mertsj):

We know the vertices so we can find a

OpenStudy (mertsj):

Yes. a = 3, = 4

OpenStudy (mertsj):

b=4, I meant

OpenStudy (anonymous):

Ok, I got that from the asymptote

OpenStudy (mertsj):

Very good.

OpenStudy (anonymous):

So this is a vertical hyperbola

OpenStudy (mertsj):

yes

OpenStudy (anonymous):

So is the equation y^2/9 - x^2/16 = 1 ?

OpenStudy (mertsj):

We have to be careful here because from the asymptotes, it looks like a = 3 and b = 4 but it tells us the vertices are (0,6) and (0,-6) so we know that a is really 6 and the equations of the asymptotes before they reduced the fraction were y =( 6/8)x and y = (-6/8)x

OpenStudy (anonymous):

Ohhhh

OpenStudy (anonymous):

I knew something was weird

OpenStudy (mertsj):

So the equation is?

OpenStudy (anonymous):

So it's y^2/36 - x^2/64 = 1 ?

OpenStudy (mertsj):

Perfect!!

OpenStudy (anonymous):

Thanks!! You rock!

OpenStudy (mertsj):

ty and yw

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