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BeccaB003
 one year ago
Check my answer please? Thanks!!
BeccaB003
 one year ago
Check my answer please? Thanks!!

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BeccaB003
 one year ago
Best ResponseYou've already chosen the best response.1Find the slopes of the asymptotes of a hyperbola with the equation y^2 = 36 + 4x^2 I got 1/4 and 1/4

BeccaB003
 one year ago
Best ResponseYou've already chosen the best response.1Then how do you solve? I can show you my steps if you'd like.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1sure go ahead and post them please

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'll point out where you went wrong and then show how to do it

BeccaB003
 one year ago
Best ResponseYou've already chosen the best response.1okay, original equation: \[y^2 = 36 + 4x^2\] I put it in standard form: \[y^24x^2=36\] Then divide by 36 to get everything to equal 1 \[\frac{ y^24x^2=36 }{ 36 }\] \[\frac{ y^2 }{ 36 }  \frac{ x^2 }{ 9 }=1\] Slopes = \[\frac{ b }{ a }\] b = 3 a = 6 \[\frac{ 3 }{ 6 } = \frac{ 1 }{ 2 }\]

BeccaB003
 one year ago
Best ResponseYou've already chosen the best response.1hmm... 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
 one year ago
Best ResponseYou've already chosen the best response.1If you look at this page https://www.purplemath.com/modules/hyperbola.htm you'll notice what I'm attaching as an image

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1a^2 = 36 so a = 6 b^2 = 9 so b = 3 slope = +a/b = +6/3 = +2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1so you were very close when you got 1/2 and 1/2 you just had it flipped

BeccaB003
 one year ago
Best ResponseYou've already chosen the best response.1Wait, I thought the equation for slopes was b/a not a/b.... hmm... thanks for your help!! :D

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1it 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
 one year ago
Best ResponseYou've already chosen the best response.1for some reason, the 'a' and 'b' switch places the 'b' should be with the (yk) term. I'm not sure why it went with the (xh) term

BeccaB003
 one year ago
Best ResponseYou've already chosen the best response.1Oh! I see!! I always forget about how what equation you use is dependent on how the hyperbola is oriented.

BeccaB003
 one year ago
Best ResponseYou've already chosen the best response.1So, if the equations was going a different direction on the graph it would be b/a not a/b?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah if the hyperbola is opening left/right, then you use b/a as the slope

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you're welcome
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