Find the vertices and foci of the hyperbola with equation x squared over sixteen minus y squared over forty eight = 1

Vertices: (0, ± 4); Foci: (0, ±8)
Vertices: (0, ± 8); Foci: (0, ± 4)
Vertices: (± 4, 0); Foci: (± 8, 0)
Vertices: (± 8, 0); Foci: (± 4, 0)

Question

Find the vertices and foci of the hyperbola with equation x squared over sixteen minus y squared over forty eight = 1

 Vertices: (0, ± 4); Foci: (0, ±8)
 Vertices: (0, ± 8); Foci: (0, ± 4)
 Vertices: (± 4, 0); Foci: (± 8, 0)
 Vertices: (± 8, 0); Foci: (± 4, 0)

anonymous · Answer

cancel out A and B cuz foci are always subtracted from x and y. (That's how it was with parabola). I'm not sure...

directrix · Answer

If you are allowed to use technology, go to this link and scroll down to "Hyperbola Properties." http://www.wolframalpha.com/input/?i=Foci+%28x%5E2%29%2F16+-%28+y%5E2%29%2F48+%3D+1

sarahc · Answer

attachment

sarahc · Answer

@Directrix

directrix · Answer

From @Cyber_Friend_David

Draw the two asy. Now draw the two points. Draw two horizontal lines through the points until the asymptotes. Where they intersect is the important information. It should be (8,6) with a combination of +/-'s. because the points lie on the y axis, the y term is positive. 
 The values (8,6) are the denominators for the x and y terms respectively 
 Y^2/6^2 - x^2/8^2=1 
 Y^2/36-x^2/64=1