SOLUTION: Find the slopes of the asymptotes of a hyperbola with the equation y² = 36 + 4x².

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the slopes of the asymptotes of a hyperbola with the equation y² = 36 + 4x².      Log On


   



Question 589278: Find the slopes of the asymptotes of a hyperbola with the equation
y² = 36 + 4x².

Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
y² = 36 + 4x².
We have to get it in either the form:

x%5E2%2Fa%5E2 - y%5E2%2Fb%5E2 = 1

which looks like this  ) (
and has slopes of asymptotes %22%22+%2B-+b%2Fa

or the form:

y%5E2%2Fa%5E2 - x%5E2%2Fb%5E2 = 1

which looks like this 

and has slopes of asymptotes %22%22+%2B-+a%2Fb

      y² = 36 + 4x²

y² - 4x² = 36

Get 1 on the right by dividing through by 36

y%5E2%2F36 + 4x%5E2%2F36 = 36%2F36

y%5E2%2F36 + x%5E2%2F9 = 1

y%5E2%2F6%5E2 + x%5E2%2F3%5E2 = 1


So, it's in the form

y%5E2%2Fa%5E2 - x%5E2%2Fb%5E2 = 1, with a=6, and b=3

which looks like this 

and has slopes of asymptotes %22%22+%2B-+a%2Fb = %22%22+%2B-+6%2F3 = ±2

Edwin