Answer
$(0,\pm2)$, $(0,\pm\sqrt 5)$; graph (a) .
Work Step by Step
From the equation, we have
$a=2, b=1, c=\sqrt {2^2+1^2}=\sqrt 5$
with a vertical transverse axis. Thus the vertices are at $(0,\pm2)$ and the foci are at $(0,\pm\sqrt 5)$. Only graph (a) fits the parameters.
