Answer
See below:
Work Step by Step
The standard equation of a hyperbola with center $(h,k)$ and horizontal transverse axis is
$\frac{{{\left( x-a \right)}^{2}}}{{{a}^{2}}}-\frac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$
Consider the equation ${{x}^{2}}-4{{y}^{2}}-4x-24y-48=0$
Now subtract/add $4$ in the equation and rearrange the terms:
${{x}^{2}}-4{{y}^{2}}-4x-24y-48+4-4=0$
Re-arrange the terms
$\left( {{x}^{2}}+4-4x \right)-4{{y}^{2}}-24y-48-4=0$
Now subtract $36$ in the equation and rearrange the terms, $\begin{align}
& \left( {{x}^{2}}+4-4x \right)-4{{y}^{2}}-24y-36+36-48-4=0 \\
& \left( {{x}^{2}}+4-4x \right)+\left( -4{{y}^{2}}-24y-36 \right)+36-48-4=0 \\
& \left( {{x}^{2}}+4-4x \right)+\left( -4{{y}^{2}}-24y-36 \right)-16=0
\end{align}$
Take out $-4$ and complete the whole square.
$\begin{align}
& {{\left( x-2 \right)}^{2}}+-4{{\left( {{y}^{2}}+6y+9 \right)}^{2}}=16 \\
& {{\left( x-2 \right)}^{2}}-4{{\left( y+3 \right)}^{2}}=16
\end{align}$
Divide both sides of the equation by $16$
$\begin{align}
& \frac{{{\left( x-2 \right)}^{2}}-4{{\left( y+3 \right)}^{2}}}{16}=1 \\
& \frac{{{\left( x-2 \right)}^{2}}}{16}-\frac{{{\left( y+3 \right)}^{2}}}{4}=1 \\
\end{align}$
The last expression can be compared to the general equation of a hyperbola with horizontal transverse axis $\frac{{{\left( x-a \right)}^{2}}}{{{a}^{2}}}-\frac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$ and center $\left( h,k \right)$.
Thus, $\frac{{{\left( x-2 \right)}^{2}}}{{{4}^{2}}}-\frac{{{\left( y+3 \right)}^{2}}}{{{2}^{2}}}=1$ with center $\left( 2,-3 \right)$.
To plot the graph, asymptotes are required to be calculated.
Asymptotes are given by $ y=k\pm \frac{b}{a}\left( x-h \right)$
For the hyperbola above, the asymptotes are $\begin{align}
& y=-3\pm \frac{2}{4}\left( x-2 \right) \\
& y+3=\pm \frac{1}{2}\left( x-2 \right) \\
\end{align}$
Which gives two asymptotes with equation $ y+3=\frac{1}{2}\left( x-2 \right)$ and $ y+3=-\frac{1}{2}\left( x-2 \right)$
Plot these asymptotes.
First consider $ y+3=\frac{1}{2}\left( x-2 \right)$
This can be written as $\begin{align}
& y+3=\frac{1}{2}\left( x-2 \right) \\
& y=0.5x-1-3 \\
& y=0.5x-4 \\
\end{align}$.
This is the same as the graph of the line $ y=x $ but with two transformations.
First, horizontally shrink by $0.5$ units and then vertically shift downwards by $4$ units.
Now, consider $ y+3=-\frac{1}{2}\left( x-2 \right)$
This can be written as
$\begin{align}
& y+3=-\frac{1}{2}\left( x-2 \right) \\
& y+3=-0.5x+1 \\
& y=-0.5x-2
\end{align}$.
This is the same as the graph of the line $ y=x $ but with two transformations.
We horizontally shrink by $-0.5$ units and then vertically shift downwards by $2$ units.
