# 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. $x^2 - 4x - 4 - 2y^2 - 4y = 0\to (x-2)^2 -2(y+1)^2 =6 \to \dfrac{(x-2)^2}{(\sqrt{6})^2}-\dfrac{(y+1)^2}{(\sqrt{3})^2}=1\to a = \sqrt{6}, b = \sqrt{3}\to c = \sqrt{6+3} = 3\to F= (2\pm 3,-1)=(-1,-1), (5,-1)\to \text{ distance to x-axix = 1}$
$$\frac{(x + 2)^2}{2} - \frac{(y - 1)^2}{1}; C = (-2, 1)$$
$$C = \sqrt{2 + 1} = \sqrt{3}$$
Answer: $$1 + \sqrt(3) = 2.73$$