BeccaB003
  • BeccaB003
Check my answer please? Thanks!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
BeccaB003
  • BeccaB003
Find the slopes of the asymptotes of a hyperbola with the equation y^2 = 36 + 4x^2 I got 1/4 and -1/4
jim_thompson5910
  • jim_thompson5910
incorrect
BeccaB003
  • BeccaB003
Then how do you solve? I can show you my steps if you'd like.

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jim_thompson5910
  • jim_thompson5910
sure go ahead and post them please
jim_thompson5910
  • jim_thompson5910
I'll point out where you went wrong and then show how to do it
BeccaB003
  • BeccaB003
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 \[\frac{ 3 }{ 6 } = \frac{ 1 }{ 2 }\]
BeccaB003
  • BeccaB003
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...
jim_thompson5910
  • jim_thompson5910
If you look at this page https://www.purplemath.com/modules/hyperbola.htm you'll notice what I'm attaching as an image
jim_thompson5910
  • jim_thompson5910
a^2 = 36 so a = 6 b^2 = 9 so b = 3 slope = +-a/b = +-6/3 = +-2
jim_thompson5910
  • jim_thompson5910
so you were very close when you got 1/2 and -1/2 you just had it flipped
BeccaB003
  • BeccaB003
Wait, I thought the equation for slopes was b/a not a/b.... hmm... thanks for your help!! :D
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
Oh I see now
jim_thompson5910
  • jim_thompson5910
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
BeccaB003
  • BeccaB003
Oh! I see!! I always forget about how what equation you use is dependent on how the hyperbola is oriented.
BeccaB003
  • BeccaB003
So, if the equations was going a different direction on the graph it would be b/a not a/b?
jim_thompson5910
  • jim_thompson5910
yeah if the hyperbola is opening left/right, then you use b/a as the slope
BeccaB003
  • BeccaB003
Thanks!
jim_thompson5910
  • jim_thompson5910
you're welcome

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