Answer
$ a.\qquad$
New center:$\quad(0, -2)$
New foci: $\quad (0,-5)$ and $(0, 1)$
New vertices: $\quad (0, -4)$ and $(0,0)$
New asymptotes,$\quad y= \displaystyle \pm\frac{2}{\sqrt{5}}x-2$
$ b.\qquad$
See graph.
Work Step by Step
$ a.\qquad$
$\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{5}=1\quad $
Which has foci on the y-axis, $ a=2, b=\sqrt{5}$
The center-to-focus distance is:
$\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{4+5}=3$
Foci: $\quad (0,\pm c)=\quad (0,\pm 3)$
Vertices: $\quad (0,\pm a)=\quad (0,\pm 2)$
Asymptotes: $\quad$ $\displaystyle \quad y=\pm\frac{a}{b}x\Rightarrow \displaystyle \quad y=\pm\frac{2}{\sqrt{5}}x$
The translations are such that $(x,y)\rightarrow(x',y')$, where
$\left\{\begin{array}{ll}
x'=x & \text{... no shift}\\
y'=y-2 & \text{... shift down}
\end{array}\right.$
New center: $(0, -2+0)=(0, -2)$
New foci: $\quad(0,-2\pm 3)\Rightarrow \quad (0,-5)$ and $(0, 1)$
New vertices: $\quad (0,-2\pm 2)\Rightarrow \quad (0, -4)$ and $(0,0)$
New asymptotes are,
$\quad y'=\displaystyle \pm\frac{2}{\sqrt{5}}x', \quad \left\{\begin{array}{lll}
x'=x & ... & x=x'\\
y'=y-2 & & y=y'+2
\end{array}\right.$
$(y+2)= \displaystyle \pm\frac{2}{\sqrt{5}}x$
$y= \displaystyle \pm\frac{2}{\sqrt{5}}x-2$
$ b.\qquad$
See graph.
