Question

The graph of (x/2)^2-(y/3)^2=1 is a hyperbola. Which set of equations represents the asymptotes of the hyperbola's graph?
A) y=(3/2)x, y=(-3/2)x
B) y=(2/3)x, y=(-2/3)x
C) y=(1/2)x, y=(-1/2)x
D) y=(1/3)x, y=(-1/3)x

Answer 1

please help(:

Answer 2

anyone?

Answer 3

A hyperbola has an equation (x-h)^2/a^2 + (y-k)^2/b^2 = 1 where h, k are constants and a, b are also constants. So now that you know a, b, h, k you can use the formula for the asymptote:

\[y = \pm \frac{ a }{ b } (x - h) + k\]

Answer 4

So can you identify the variables:

a = ?
b = ?
h = ?
k = ?

Answer 5

ive never done this before so I dont know

Answer 6

i could take a guess.
a=2
b=3
ummm

Answer 7

i have no clue

Answer 8

Well the formula you provided is:

\[\frac{ x ^{2} }{ 4 } - \frac{ y ^{2} }{ 9 } = 1\]

And the formula for a hyperbole is:

\[\frac{ (x - h)^{2} }{ a^{2} } + \frac{ (y - k)^{2} }{ b^{2} } = 1\]

Answer 9

Do you see the similarities there?

Answer 10

umm

Answer 11

hyperbole?

Answer 12

Hyperbola* sorry. Also, the formula should say - not +.

But see:
\[(x - h)^{2} = x^{2}\]
\[(y - k)^{2} = y^{2}\]
\[a^{2} = 4\]
\[b^{2} = 9\]

Answer 13

ok

Answer 14

So what are a, b, h and k?

Answer 15

a=2
b=3
h=x/2
k=y/3

Answer 16

Whoa, h and k are constants. They do not contain x or y in any way.

If (x - h)^2 = x^2, what is h?

Answer 17

0?

Answer 18

Yes. Now take your values and plug them in to the asymptote equation:

\[y = \pm \frac{ a }{ b } (x - h) + k\]

Answer 19

did we figure out wht h and k were?

Answer 20

You found h, now apply the same reasoning to find k.

Answer 21

so
\[\pm2/3(x-0)+k\]
ummm idk

Answer 22

oh k is 0..my bad

Answer 23

What is the correct answer to the problem?

Answer 24

B?

Answer 25

That is correct! Good work. Bookmark this thread if you have trouble in future with asymptotes.

Answer 26

alright. thank you so much