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

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Answer 1 · 2015-07-05T06:21:39

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

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)

How do you find the center and vertex for 9x^2 – y^2 – 36x + 4y + 23 = 09x^2 – y^2 – 36x + 4y + 23 = 0?