Question

Convert the equation to the standard form for a hyperbola by completing the square on x and y.

x2 - y2 + 6x - 4y + 4 = 0

Answer 1

(x^2)-(y^2) + 6x -4y +4 = 0
(x^2) + 6x -(y^2) - 4y + 13 = 9
((x+3)^2) -(y^2) - 4y +4 = 9
-((x+3)^2) + (y^2) + 4y -4 = -9
-((x+3)^2) + (y^2) + 4y = -5
-((x+3)^2) + (y^2) + 4y +4 = -1
-((x+3)^2) + ((y+2)^2) = -1
((x+3)^2) - ((y+2)^2) = 1
(x + 3)2 - (y + 2)2 = 1
is this right