# Convert the equation to the standard form for a hyperbola by completing the square on x and y. x2 - y2 + 6x - 4y + 4 = 0?

(x + 3)2 + (y + 2)2 = 1

(x + 3)2 - (y + 2)2 = 1

(y + 3)2- (x + 2)2 = 1

(y+3)^2/16 − (x+2)^2/36 = 1

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tramserranX² + 6x + ____ - (y² + 4y + ____ ) = -4 + ____ + ____ x² + 6x + 9 - (y² + 4y + 4 ) = -4 + 9 + -4 (x + 3)² - (y + 2)² = 1 Answer: B