Question 38908
From
9x^2 - 4y^2 - 16y - 52 = 0
we complete the square on the y's...add 52 to both sides too...
9x^2 - 4y^2 - 16y - 16 = 52 - 16
9x^2 - 4(y - 2)^2 = 36
divide by 36 and get
x^2 / 4 - (y - 2)^2 / 9  = 1
This is a hyperbola, lobes left and right, centered at (0, 2) with asymptotes of slope ± 3/2