x2−24y2−12x+96y−96=0. Find center,vertices,foci,and equation of asymptotes

Question

It's a hyperbola.

I

Answer 1

We first need to get this into the standard form for a hyperbola:

x² / a² − y² / b²=1 (Assumes a "horizontal" hyperbola, with the transverse axis parallel to the x-axis)

To do this, we will "complete the square" for the x and y terms:

x² − 24y² − 12x + 96y − 96 = 0

Rearrange:

x² − 12x __________− 24*(y² - 4y __________) = 96

Add constants to create perfect squares:

x² − 12x + 36 − 24*(y² - 4y + 4) = 96 +36 - 24(4) = 36

Put into the standard format:

(x - 6)² - 24*(y - 2)² = 36

(x - 6)² / 36 - (y - 2)² / 1.5 = 1

The offsets tell us that the center is at (+6,-2). Use the hyperbola definitions to find the other parameters