Question

Find the standard form of the equation of the hyperbola with the given characteristics.
foci: (+-3,0)
Asymptotes: y=+-2x

Answer 1

since the foci are located parallel to the x axis; we can make a note of that and say that they open to the left and right

Answer 2

I know the origin is at (0,0)
my equation should look like: y^2/a^2-x^2/b^2=1

Answer 3

the center appears to already be at 0; otherwise youd have different 'x' parts for the foci

Answer 4

jk it would be
x^2/a^2-y^2/b^2=1

Answer 5

the asymptote tells us what a and b are really; since it is in the form y = (b/a)x

Answer 6

2 b
-- x = --- x
1 a

the y component is b and the x compnent is a

Answer 7

Right and I know I need to find c^2=a^2+b^2
I know C is 3^2 right?

Answer 8

x^2 y^2
--- - --- = 1
1^2 2^2

Answer 9

a is only needed if you are looking to find a or b; we know both a and b from the asympotote

Answer 10

ok but that can't be right....I think we may have a wrong....

Answer 11

my options are
x^2 y^2
--- - --- = 1
9 4

x^2 y^2
--- - --- = 1
9/5 36/5

x^2 y^2
--- - --- = 1
36/5 9/5

Answer 12

x^2 y^2
--- - --- = 1
9 4

I would guess this one but I'm not sure

Answer 13

lets recheck then :)

Answer 14

we need to determine the "rectangle" of the hyperbola; to so that, lets write it as an ellipst

Answer 15

uh.. ok?

Answer 16

x^2 y^2
--- + ---- = 1 the vertices of this ellipse are the top and sides
a^2 b^2 of your asympotoe information

Answer 17

correct

Answer 18

2x is the asympy.... this tells us the from the center, we go up 2 and over 1 to for the top right corner of the rectangle; and our ellipse will be 2 high and 1 left right?

Answer 19

right

Answer 20

ok I got it

Answer 21

and it should end up looking like this inthe end

Answer 22

ok.... ugh this stupid math class their answer key never matches up to what I get lol

Answer 23

our y compnent is 2^2 regardless right :)

Answer 24

id go with your first option;

dbl chk by making y=0 and seeing what you get for an x value.. maybe :)

Answer 25

ok thanks