Answer
Center: $(0,3)$
Foci: $(0,0)$ and $(0,6)$
Vertices: $(0, 3\pm \sqrt 6)$
Asymptote: $y=±\sqrt 2x+3$
Work Step by Step
$2x^2−y^2+6y=3$
$2x^2−(y^2-6y + \textbf{9})=3 \textbf{ - 9}$
$2x^2 - (y-3)^2=-6$
Hyperbola form: $\frac{(y-3)^2}{6} - \frac{x^2}{3}=1$
$\qquad \qquad \qquad \enspace \frac{(y-3)^2}{(\sqrt 6)^2} - \frac{x^2}{(\sqrt 3)^2}=1$
$\qquad \qquad \qquad \enspace \qquad \enspace h=0,k=3,a=\sqrt 3,b=\sqrt 6$
Linear eccentricity (focal distance): $c= \sqrt{a^2+b^2}$
$\qquad \qquad \qquad \qquad \qquad \hspace{3em} c=\sqrt{(\sqrt3)^2+(\sqrt6)^2}$
$\qquad \qquad \qquad \qquad \qquad \hspace{3em} c=3$
Center: $(h,k)$
$\qquad \quad (0,3)$
Foci: $(h,k±c)$
$\qquad (0,3±3)$
$\qquad (0,0)$ and $(0,6)$
Vertices: $(h,k±b)$
$\qquad \quad \enspace (0 , 3\pm \sqrt 6)$
$\qquad \quad \enspace (0, 3\pm \sqrt 6)$
Asymptote: $y=±\frac{b}{a}(x−h)+k$
$\qquad \qquad \enspace y=±\frac{\sqrt 6}{\sqrt 3}(x-0)+3$
$\qquad \qquad \enspace y=±\sqrt 2x+3$
