Question

How do you find the center and vertex for `9x^2 – y^2 – 36x + 4y + 23 = 0`?

Answer 1

center (2,2)
vertices would be (1,2) and (3,2)

Explanation:

First, we can complete the squares of x and y, by writing it as follows:
`9x^2 -36x- (y^2 -4y)= -23`, or

`9(x^2 -4x +4)- (y^2 -4y +4)= 36-23-4` or,

`9(x^2 -4x+4)-(y-2)^2=9`, or,

`9(x-2)^2 - (y-2)^2 =9` or,

`(x-2)^2 /1 - (y-2)^2 /9 =1`

This equation represents a hyperbola with its center at (2,2). Its vertices would be (1,2) and (3,2)