Question

Find the vertices and foci of the hyperbola with equation x^2/16- y^2/48=1

Answer 1

Just a sec buddy, writing it out.
Meanwhile, this link might help you learn more:
http://www.purplemath.com/modules/hyperbola2.htm

Answer 2

thank you i'll read that info meanwhile!

Answer 3

Just a note
The vertices are (h-a, k) and (h+a, k)
The foci are (h-c, k) and (h+c, k)

Answer 4

\[\large \frac{x^2}{16}-\frac{y^2}{48}=1\]

foci: \[\large c^2 = a^2 - b^2\]\[a^2 = 16 \ \ b^2 = 48 \]\[c^2 = 16 + 48\]\[c=\sqrt{16+48} = \sqrt{ 64} = 8\]
Foci = \(\large \pm 8\)

Answer 5

Sorry, we know the center is at (0,0) and as @bahrom7893 stated, our foci follow the equation (h-c, k) and (h+c, k) So we have -8 and +8, inputting them into this formula, we would have \[(h+c,k) = (0 + 8, 0) = (8,0) \\ (h-c,k) = (0-8,0) = (-8,0)\]

Answer 6

And now our vertices....

Answer 7

so my final answer is going to be
Vertices: (± 4, 0); Foci: (± 8, 0) ???

Answer 8

Yeah, believe so.

Answer 9

Thanks you guys.

Answer 10

thank you so much!