Answer
The centre is $\left( 4,-3 \right)$
The vertices are at $\left( 4,-1 \right)\text{ and }\left( 4,-5 \right)$
The asymptotes are $\left( y+3 \right)=\pm \frac{1}{2}\left( x-4 \right)$
The graph is shown below.
Work Step by Step
$8{{\left( y+3 \right)}^{2}}-2{{\left( x-4 \right)}^{2}}=32$
Divide by $32$
$\frac{{{\left( y+3 \right)}^{2}}}{4}-\frac{{{\left( x-4 \right)}^{2}}}{16}=1$
$\frac{{{\left( y-\left( -3 \right) \right)}^{2}}}{{{2}^{2}}}-\frac{{{\left( x-4 \right)}^{2}}}{{{4}^{2}}}=1$.
Compare this equation with the standard equation$\frac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}-\frac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}=1$,
$h=4,k=-3,a=4,b=2$
Thus, the centre of the hyperbola $\left( h,k \right)$ is $\left( 4,-3 \right)$.
The vertices of the hyperbola are,
$\left( h,k+b \right)$ and $\left( h,k-b \right)$
Now, put $h=4$, $k=-3$ and $b=2$ in the first vertices,
$\left( 4,-3+2 \right)=\left( 4,-1 \right)$
Now, put $h=4$, $k=-3$ and $b=2$ in the second vertices,
$\left( 4,-3-2 \right)=\left( 4,-5 \right)$
Hence, the vertices of the hyperbola $\left( h,k+b \right)$ and $\left( h,k-b \right)$ are $\left( 4,-1 \right)$ and $\left( 4,-5 \right)$.
The equation of the asymptote for the standard equation of a hyperbola is,
$\left( y-k \right)=\pm \frac{b}{a}\left( x-h \right)$
Now, put $h=4$, $k=-3$, $a=4$, $b=2$ in the equation.
Thus, the equations of the asymptotes are $\left( y+3 \right)=\frac{1}{2}\left( x-4 \right)$ and $\left( y+3 \right)=-\frac{1}{2}\left( x-4 \right)$.
Graph:
Consider the equation of the hyperbola,
$\frac{{{\left( y-\left( -3 \right) \right)}^{2}}}{{{2}^{2}}}-\frac{{{\left( x-4 \right)}^{2}}}{{{4}^{2}}}=1$
First, graph the asymptotes.
Then, plot points $\left( 8,-1 \right)$, $\left( 8,-5 \right)$, $\left( 0,-1 \right)$, $\left( 0,-5 \right)$ and lightly sketch a rectangle.
Draw curves through the vertices toward the asymptotes.
The graph of the hyperbola $\frac{{{\left( y+3 \right)}^{2}}}{4}-\frac{{{\left( x-4 \right)}^{2}}}{16}=1$ is shown below:
