1. 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

2. jim_thompson5910

incorrect

3. BeccaB003

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

4. jim_thompson5910

5. jim_thompson5910

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

6. 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 $\frac{ 3 }{ 6 } = \frac{ 1 }{ 2 }$

7. 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...

8. 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

9. jim_thompson5910

a^2 = 36 so a = 6 b^2 = 9 so b = 3 slope = +-a/b = +-6/3 = +-2

10. jim_thompson5910

so you were very close when you got 1/2 and -1/2 you just had it flipped

11. BeccaB003

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

12. 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

13. jim_thompson5910

Oh I see now

14. 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

15. BeccaB003

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

16. BeccaB003

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

17. jim_thompson5910

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

18. BeccaB003

Thanks!

19. jim_thompson5910

you're welcome