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