Question

Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ±1/4x.

Answer 1

http://2.bp.blogspot.com/-PA3nK9JlU1o/Td8gGQLPRTI/AAAAAAAAACE/mTAESwctzio/s320/hyperbola.jpg

Answer 2

Jdoe! Thanks lol

Answer 3

notice, your vertices are at (0, 4) and (0, -4)
the "x" doesn't change, the "y" coordinate does
that means the hyperbola is moving upward/downward

Answer 4

the distance between one vertex to the other is 4 units up plus 4 units down, so 8 units, so half-way in between, is the center, from (0, 4) to (0, -4), puts you in the origin
moving upwards/downwads by 4 units to the vertices, means, THAT IS the TRAVERSE AXIS, and thus the "a" component of the hyperbola
so a = 4
center = (0, 0)

Answer 5

since the traverse axis is the "y" axis, that means the fraction with the POSITIVE sign will be the fraction with the "y", so thus far it'd look like
$$
\cfrac{(y-0)^2}{4}-\cfrac{(x-0)^2}{b} = 1
$$

Answer 6

Hmmm okay i see

Answer 7

y^2/16 - x^2/64 = 1

Answer 8

a = 4
\(a^2 = 2^2\)

Answer 9

so the 16 is correct

Answer 10

wait.. that made no sense lol

Answer 11

lololololl

Answer 12

right it'd be 16 hehe

Answer 13

lol so what do i dooo? if a = 4 then a^2 = 16, meaning that the 16 is right

Answer 14

is the 64 right?

Answer 15

or should it be 256?:

Answer 16

well, dunno thus far, you see the asymptotes for a vertical hyperbola are at
\(y = k \pm \cfrac{a}{b}(x-h)\\\)

Answer 17

:l i think im just gonna guess... been on this one Q for 35 mins

Answer 18

which leads me to think b is 1

Answer 19

if he slope of the asymptote is 1/4, and a = 4, then b has to be the numerator