SOLUTION: Hello, could someone please help me with the following problem? I am supposed to provide the equation of the asymptotes for the hyperbola: 4y^2-x^2=36. Thank you for your assist

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Hello, could someone please help me with the following problem? I am supposed to provide the equation of the asymptotes for the hyperbola: 4y^2-x^2=36. Thank you for your assist      Log On


   

Question 1088558: Hello, could someone please help me with the following problem? I am supposed to provide the equation of the asymptotes for the hyperbola:
4y^2-x^2=36. Thank you for your assistance!
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
4y%5E2-x%5E2=36 in standard form %28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1, we have
4y%5E2%2F36-x%5E2%2F36=36%2F36
y%5E2%2F9-x%5E2%2F36=36%2F36
y%5E2%2F3%5E2-x%5E2%2F6%5E2=1=> h=0,k=0,a=3, and b=6

foci: (0, -3sqrt%285%29) and (0, 3sqrt%285%29)≈(0, -6.7) and (0, 6.7)
vertices:(0, -3) and (0, 3)
center: (0, 0)
semi-major axis length:+3
semi-minor axis length:6

asymptotes equation: y= ± %28a%2Fb%29%28x-k%29%2Bh; since h=0,k=0,a=3, and b=6
we have:
y= ± %283%2F6%29%28x-0%29%2B0
y+=+x%2F2+ or y+=+-x%2F2