Step 1:
$x^2-3y^2+6x+6y+18=0$
$x^2-3y^2+6x+6y=-18$
Completing the squares,
$(x^2+6x+9)-3(y^2-2y+1)=-18+9-3$
$(x+3)^2-3(y-1)^2=-12$
The above equation is divided by $-12$ we get
$\large\frac{(y-1)^2}{4}-\large\frac{(x+3)^2}{12}$$=1$
$a^2=4,b^2=12$
$\Rightarrow a=2,b=2\sqrt 3$
Step 2:
Shifting the origin to $(-1,3)$ by translation of axes.
$X=x+3$,$Y=y-1$
The equation reduces to the form $\large\frac{Y^2}{4}-\large\frac{X^2}{12}$$=1$
The $Y$-axis is the transverse axis.
Step 3:
$XY$-axes :
$e=\sqrt{1+\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1+\large\frac{12}{4}}=$$\sqrt4=2$
Centre :$(0,0)$
Foci : $(0,\pm ae)$
$\Rightarrow (0,\pm 2\times 2)$
$\Rightarrow (0,\pm 4)$
Vertices : $(0,\pm a)$
$\Rightarrow (0,\pm 2)$
Step 4:
Centre : $(1,-3)$
Foci : $(1,-3\pm 4)$
(i.e) $(1,7),(1,1)$
Vertices : $(1,-3\pm 2)$
(i.e) $(1,-5),(1,-1)$