Answer
$(0,0)$
Vertices:$(h \pm a, k) =(\pm 1, 0)$
Foci: $(h \pm c, k) =(\pm \sqrt 5, 0)$
Work Step by Step
The standard form of the equation of the hyperbola with a horizontal transverse axis can be expressed as: $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
The vertices and foci have the form $(\pm a, 0) $ and $(\pm c,0)$.
The standard form of the equation of the hyperbola with a vertical
transverse axis can be expressed as: $\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
The vertices and foci have the form $(0, \pm, a) $ and $(0, \pm c)$.
The center is the midpoint of the vertices: $(0,0)$
We have: $ a=1; b=2$
$c=\sqrt {a^2+b^2}=\sqrt {1^2+2^2}=\sqrt 5$
and $\dfrac{x^2}{1^2}-\dfrac{y^2}{2^2}=1$
Vertices: $(h \pm a, k) =(\pm 1, 0)$
Foci: $(h \pm c, k) =(\pm \sqrt 5, 0)$
