Answer
Same center, different vertices, different foci
Work Step by Step
We are given the hyperbolas:
$\dfrac{x^2}{9}-\dfrac{y^2}{1}=1$
$\dfrac{y^2}{9}-\dfrac{x^2}{1}=1$
We notice that both hyperbolas have the same $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=9\Rightarrow a=\sqrt 9=3$
$b^2=1\Rightarrow b=\sqrt 1=1$
$c^2=a^2+b^2$
$c^2=9+1$
$c^2=10$
$c=\sqrt{10}$
Therefore the hyperbolas have the same center $(0,0)$.
Even if $a, c$ is the same, the hyperbolas have different vertices and different foci:
Vertices for the first hyperbola:
$(h-a,k)=(0-3,0)=(-3,0)$
$(h+a,k)=(0+3,0)=(3,0)$
Vertices for the second hyperbola:
$(h,k-a)=(0,0-3)=(0,-3)$
$(h,k+a)=(0,0+3)=(0,3)$
Foci for the first hyperbola:
$(h-c,k)=(0-\sqrt{10},0)=(-\sqrt{10},0)$
$(h+c,k)=(0+\sqrt{10},0)=(\sqrt{10},0)$
Foci for the second hyperbola:
$(h,k-c)=(0,0-\sqrt{10})=(0,-\sqrt{10})$
$(h,k+c)=(0,0+\sqrt{10})=(0,\sqrt{10})$
