Answer
$ a.\qquad$
New center:$\quad(2,0)$
New foci: $\quad (-3,0)$ and $(7,0)$
New vertices: $\quad (-2,0)$ and $(6,0)$
New asymptotes,$\quad y= \displaystyle \pm\frac{3}{4}(x-2)$
$ b.\qquad$
See graph.
.
Work Step by Step
$ a.\qquad$
$\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{9}=1\quad $
Which has foci on the x-axis,$ a=4, b=3$
Center-to-focus distance: $\quad c=\sqrt{a^{2}+b^{2}}=\sqrt{16+9}=5$
Foci: $\quad (\pm c, 0)=\quad (\pm 5, 0)$
Vertices: $\quad (\pm a, 0)=\quad (\pm 4, 0)$
Asymptotes: $\quad$ $\quad y=\displaystyle \pm\frac{b}{a}x\Rightarrow \displaystyle \quad y=\pm\frac{3}{4}x$
The translations are such that $(x,y)\rightarrow(x',y')$, where
$\left\{\begin{array}{ll}
x'=x+2 & \text{... shift right}\\
y'=y & \text{... no shift}
\end{array}\right.$
New center: $(2+0, 0)=(2,0)$
New foci: $\quad(2\pm 5, 0)\Rightarrow \quad (-3,0)$ and $(7,0)$
New vertices: $\quad (2\pm 4, 0)\Rightarrow \quad (-2,0)$ and $(6,0)$
New asymptotes are:
$\quad y'=\displaystyle \pm\frac{3}{4}x', \quad \left\{\begin{array}{cc}
x'=x+2 ,& x=x'-2\\
y'=y ,& y=y'
\end{array}\right.$
$y= \displaystyle \pm\frac{3}{4}(x-2)$
$ b.\qquad$
See graph.
