Answer
$ \displaystyle \frac{(x-2)^{2}}{4}-\frac{(y-2)^{2}}{5}=1$
New foci:$\quad (5, 2)$ and $(-1,2)$
New vertices:$\quad (4, 2)$ and $(0,2)$
New center:$\quad (2,2)$
New asymptotes:$\displaystyle \quad y-2=\pm\frac{\sqrt{5}}{2}(x-2) $
Work Step by Step
Shift right by $2 \Rightarrow$ in the equation, replace $x$ with $x-2$
Shift up by $2 \Rightarrow$ in the equation, replace $y$ with $y-2$
New equation:$\displaystyle \quad \frac{(x-2)^{2}}{4}-\frac{(y-2)^{2}}{5}=1$
The hyperbola $\displaystyle \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$ has the form for which:
foci are on the x-axis$, \quad a=2, b=\sqrt{5},$
Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{4+5}=3$
Foci: $\quad (\pm c, 0)= \quad (\pm 3, 0)$
Vertices: $\quad (\pm 2, 0)$
Asymptotes:$\quad y=\displaystyle \pm\frac{b}{a}x\Rightarrow \quad y=\displaystyle \pm\frac{\sqrt{5}}{2}x$
The translations are such that $(x,y)\rightarrow(x',y')$, where
$\left\{\begin{array}{ll}
x'=x+2 & \text{... shift right}\\
y'=y+2 & \text{... shift up}
\end{array}\right.$
New foci:$\quad (\pm 3+2, 0+2)\Rightarrow\quad (5, 2)$ and $(-1,2)$
New vertices:$\quad (\pm 2+2, 0+2)\Rightarrow\quad (4, 2)$ and $(0,2)$
New center:$\quad (0+2,0+20)\quad =(2,2)$
New asymptotes:$\displaystyle \quad y'=\pm\frac{\sqrt{5}}{2}x'$
$y-2=\displaystyle \pm\frac{\sqrt{5}}{2}(x-2)$
