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

Jul 5, 2015

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:
$9 {x}^{2} - 36 x - \left({y}^{2} - 4 y\right) = - 23$, or

$9 \left({x}^{2} - 4 x + 4\right) - \left({y}^{2} - 4 y + 4\right) = 36 - 23 - 4$ or,

$9 \left({x}^{2} - 4 x + 4\right) - {\left(y - 2\right)}^{2} = 9$, or,

$9 {\left(x - 2\right)}^{2} - {\left(y - 2\right)}^{2} = 9$ or,

${\left(x - 2\right)}^{2} / 1 - {\left(y - 2\right)}^{2} / 9 = 1$

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