Question

pls help me with precal :(
Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at
y = ±three divided by four times x.

Answer 1

we can do this

Answer 2

ready ?

Answer 3

actually this one is not that bad
do you know what this sucker looks like ?

Answer 4

uhhhh what do you mean what it looks like? what a hyperbola with a vertice at 0,+/-6 and asymptotes at the given point look like? yea

Answer 5

ok so you know that the center is \((0,0)\) and the form of it must be
\[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\] the \(y^2\) term comes first because of the they way it is oriented

Answer 6

the fact that the vertices are at \((0,6)\) and \((0,-6)\) tells you it looks like
\[\frac{y^2}{6^2}-\frac{x^2}{b^2}=1\] and all you need is \(b\)

Answer 7

so how do i find b? should i just plug in numbers and test?

Answer 8

you have another piece of information
asymptotes at \(y=\pm\frac{3}{4}x\)

Answer 9

typo there
i meant since \(a=6\) and \(\frac{a}{b}=\frac{3}{4}\) you get \(b=8\)

Answer 10

ohhhh ok

Answer 11

and that is done now

Answer 12

Ok thanks man :^)

Answer 13

yw