SOLUTION: I cannot figure this out it's stressing me out.....I hope u can help me. PLEASE 16y^2-9x^2+36x-32y-164=0 I need to know the coordinates, and if its hyperbolic(center and focci) ci

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I cannot figure this out it's stressing me out.....I hope u can help me. PLEASE 16y^2-9x^2+36x-32y-164=0 I need to know the coordinates, and if its hyperbolic(center and focci) ci      Log On


   

Question 307008: I cannot figure this out it's stressing me out.....I hope u can help me. PLEASE
16y^2-9x^2+36x-32y-164=0 I need to know the coordinates, and if its hyperbolic(center and focci) circular (center and vertex) elliptical(center and focci) parabolic(vertex and focus). Please please explain. Thank you
Found 2 solutions by Fombitz, scott8148:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You need to complete the square in x and y.
+16y%5E2-9x%5E2%2B36x-32y-164=0
-9x%5E2%2B36x%2B16y%5E2-32y-164=0
-9%28x%5E2-4x%29%2B16%28y%5E2-2y%29=164
-9%28x%5E2-4x%2B4%29%2B16%28y%5E2-2y%2B1%29=164-36%2B16
-9%28x-2%29%5E2%2B16%28y-1%29%5E2=144
%28y-1%29%5E2%2F9-%28x-2%29%5E2%2F16=1+
Unequal coefficients for x%5E2 and y%5E2 and unequal signs are the signs of a hyperbola.
There now you can solve for center and foci.

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
this is a hyperbola (second order terms, squares, have different signs)

isolating the squares will make it more recognizable

16y^2 - 32y + 16 - 9x^2 + 36x - 36 - 144 = 0

16(y - 1)^2 - 9(x - 2)^2 = 144

{[(y - 1)^2] / 9} - {[(x - 2)^2] / 16} = 1