Question

Check my answer please? Thanks!!

Answer 1

Find the slopes of the asymptotes of a hyperbola with the equation y^2 = 36 + 4x^2
I got 1/4 and -1/4

Answer 2

incorrect

Answer 3

Then how do you solve? I can show you my steps if you'd like.

Answer 4

sure go ahead and post them please

Answer 5

I'll point out where you went wrong and then show how to do it

Answer 6

okay, original equation:

\[y^2 = 36 + 4x^2\]

I put it in standard form:

\[y^2-4x^2=36\]

Then divide by 36 to get everything to equal 1

\[\frac{ y^2-4x^2=36 }{ 36 }\]

\[\frac{ y^2 }{ 36 } - \frac{ x^2 }{ 9 }=1\]

Slopes = \[\frac{ b }{ a }\]

b = 3 a = 6

\[\frac{ 3 }{ 6 } = \frac{ 1 }{ 2 }\]

Answer 7

hmm... doing it again I got 1/2.... I'm not sure what I did differently when I was typing out my steps because before I got 1/4...

Answer 8

If you look at this page

https://www.purplemath.com/modules/hyperbola.htm

you'll notice what I'm attaching as an image

Answer 9

a^2 = 36 so a = 6

b^2 = 9 so b = 3

slope = +-a/b = +-6/3 = +-2

Answer 10

so you were very close when you got 1/2 and -1/2

you just had it flipped

Answer 11

Wait, I thought the equation for slopes was b/a not a/b.... hmm... thanks for your help!! :D

Answer 12

it depends on how the hyperbola is oriented, ie how it opens up (left/right vs top/bottom)

that purplemath page describes it better with images

Answer 13

Oh I see now

Answer 14

for some reason, the 'a' and 'b' switch places

the 'b' should be with the (y-k) term. I'm not sure why it went with the (x-h) term

Answer 15

Oh! I see!! I always forget about how what equation you use is dependent on how the hyperbola is oriented.

Answer 16

So, if the equations was going a different direction on the graph it would be b/a not a/b?

Answer 17

yeah if the hyperbola is opening left/right, then you use b/a as the slope

Answer 18

Thanks!

Answer 19

you're welcome