Answer
Center: $(-2,-3)$
Foci: $(-2\pm \sqrt 2, -3)$
Vertices: $(-1, -3)$ and $(-3,-3)$
Asymptote: $y=x-1 $ and $y=-x-5$
Work Step by Step
$x^2 - y^2 +4x -6y=6$
$(x^2 +4x + \textbf{4}) -(y^2 +6y+\textbf{9})=6 \textbf {+4 - 9}$
$(x+2)^2 - (y+3)^2 = 1$
Hyperbola form: $\frac{(x+2)^2}{1^2} - \frac{(y+3)^2}{1^2} =1$
$\qquad \qquad \qquad \qquad \quad \enspace h=-2, k=-3, a=1, b=1$
Linear eccentricity (focal distance): $c=\sqrt {a^2 +b^2}$
$\qquad \qquad \qquad \qquad \qquad \qquad \quad c=\sqrt {1^2 +1^2}$
$\qquad \qquad \qquad \qquad \qquad \qquad \quad c=\sqrt {2}$
Center: $(h,k)$
$\qquad \quad (-2,-3)$
Foci: $(h \pm c, k)$
$\qquad (-2\pm \sqrt 2, -3)$
Vertices: $(h \pm a, k)$
$\qquad \quad \enspace (-2 \pm 1, -3)$
$\qquad \quad \enspace (-1, -3)$ and $(-3,-3)$
Asymptote: $y=\pm\frac{b}{a} (x-h)+k$
$\qquad \qquad \enspace y=\pm \frac{1}{1} (x+2) -3 $
$\qquad \qquad \enspace y=x-1 $ and $y=-x-5$
