Question

Find the vertices and foci of the hyperbola with equation x^2/4 - y^2/60 = 1
1) Vertices: (± 2, 0); Foci: (± 8, 0)
2) Vertices: (± 8, 0); Foci: (± 2, 0)
3) Vertices: (0, ± 8); Foci: (0, ± 2)
4) Vertices: (0, ± 2); Foci: (0, ±8)

Answer 1

for your reference:
standard form of hyperbola (horizontal axis)
(x-h)^2 / a^2 - (y-k)^2 / b^2 = 1
asymptote slope m = ± b/a

standard form of hyperbola (vertical axis)
(y-k)^2 / a^2 - (x-h)^2 / b^2 = 1
asymptote slope m = ± a/b

(h,k) => center
a: distance between center and vertices
c^2 = a^2 + b^2
c: distance between center and focus
.........................................

Answer 2

Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7).
y^2/4-x^2/45=1

Answer 3

Did that help ?

Answer 4

it did but i still don't know what i'm doing

Answer 5

what would be the center h,k

Answer 6

im not really sure thats a good question lol

Answer 7

yeah, i'm having trouble with two more problems like this

Answer 8

i can help a little

Answer 9

thats all i need

Answer 10

ok

Answer 11

so what would be the first step to solving this