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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?

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

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  • $\begingroup$ just came back to this question.... tried WA.. i still cant believe distance to x-axis is 1? wolframalpha.com/input/… $\endgroup$ – james Aug 19 '15 at 12:42
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$\frac{(x + 2)^2}{2} - \frac{(y - 1)^2}{1}; C = (-2, 1)$

$C = \sqrt{2 + 1} = \sqrt{3}$

Answer: $1 + \sqrt(3) = 2.73$

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