How do I find the asymptotes of (y-2)^2 - x^2/4 =1 The center is (0,2)
multiply by -1 to change it to "standard" form
x^2/4 - (y-2)^2/1 = -1
yes, the centre is (0,2), also
a = 2 , b = 1
Your hyperbola will have a vertical major axis, vertex at (0,3) and (0,1)
Now create a rectangular box with vertices
(4,2), (4,3), (-4,3), and (-4,1)
one asymptote will pass through (0,2) and (4,3)
slope of that one is (3-1)/(4-0) = 1/2
equation:
y = (1/2)x + b
(0,2) on it, so
2 = 0 + b
b = 2
one asymptote is y = (1/2)x + 2
find the other in the same way.