Distance of the Focus of an Hyperbola to the X-Axis For a National Board Exam Review:

How far from the $x$-axis is the focus of the hyperbola $x^2 -2y^2 + 4x + 4y + 4$?

Answer is $2.73$
Simplify into Standard Form:
$$ \frac{ (y-1)^2 }{} - \frac{ (x+2)^2 }{-2} = 1$$
$$ a^2 = 1 $$
$$ b^2 = 2 $$
$$ c^2 = 5 $$
Hyperbola is Vertical: 
$$ C(-2,1) ; y = 1 $$
$$ y = 1 + \sqrt5 = 3.24 $$
$$ y = 1 - \sqrt5 = 1.24 $$
Both answers don't match; What am I doing wrong?

Please check the formulas as you have two foci. $x^2 - 4x - 4 - 2y^2 - 4y = 0\to (x-2)^2 -2(y+1)^2 =6 \to \dfrac{(x-2)^2}{(\sqrt{6})^2}-\dfrac{(y+1)^2}{(\sqrt{3})^2}=1\to a = \sqrt{6}, b = \sqrt{3}\to c = \sqrt{6+3} = 3\to F= (2\pm 3,-1)=(-1,-1), (5,-1)\to \text{ distance to x-axix = 1}$
