Expert: Ahmed Salami - 4/25/2010

Question
I need help putting problems into Standard Form of a Hyperbola.

1. 9x^2 - 16y^2 - 36x - 64y + 116 = 0

Please Help! Thanks!

Answer
Hi Vinny,
9x^2 - 16y^2 - 36x - 64y + 116 = 0
9x^2 - 16y^2 - 36x - 64y = -116
9x^2 - 36x - 16y^2 - 64y = -116
(9x^2 - 36x) - (16y^2 + 64y) = -116
completing squares (i'm assuming you know how to),
9(x^2 - 4x) - 16(y^2 + 4y) = -116
9(x^2 - 4x + 4) - 16(y^2 + 4y + 4) = -116 + 9(4) - 16(4)
9(x - 2)^2 - 16(y + 2)^2 = -116 + 36 - 64
9(x - 2)^2 - 16(y + 2)^2 = -144
dividing through by 144,
(x - 2)^2/16 - (y + 2)^2/9 = -1
(x - 2)^2/4^2 - (y + 2)^2/3^2 = -1
OR
(y + 2)^2/3^2 - (x - 2)^2/4^2 = 1
which shows that the transverse axis is aligned with the y-axis in this case.

Regards