To help us answer the questions we will plot the vertices and the asymptotes, this is shown below:
Part I:
Since the axis of the hyperbola passes through the vertices we will have a vertical axis in this case; this means that the hyperbola is vertically oriented.
Part II:
The center of the hyperbola is the midpoint of the vertices; the midpoint between two points is given by;
[tex]M(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]Plugging the vertices (0,-6) and (0,6) on the midpoint formula we have that:
[tex]M(\frac{0+0}{2},\frac{6+(-6)}{2})=M(0,0)[/tex]Therefore, the center of the hyperbola is the point C(0,0)
Part III:
We know that the value of a is the distance from the center to the vertices; in this case, we readily notice that this distance is 6 which means that a=6. Now, we know that the asymptotes of a vertical hyperbola are given by:
[tex]y=\pm\frac{a}{b}(x-h)+k[/tex]where (h,k) is the center of the hyperbola. In this case we know h=0, k=0 and a=6 then we have:
[tex]y=\pm\frac{6}{b}x[/tex]Comparing this with the equations given we have:
[tex]\begin{gathered} \frac{6}{b}=\frac{3}{4} \\ b=\frac{6}{\frac{3}{4}} \\ b=8 \end{gathered}[/tex]Therefore, a=6 and b=8
Part IV:
The equation of a vertical hyperbola is given by:
[tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1[/tex]Plugging the values of h, k, a and b we have that:
[tex]\begin{gathered} \frac{(y-0)^2}{6^2}-\frac{(x-0)^2}{8^2}=1 \\ \frac{y^2}{36}-\frac{x^2}{64}=1 \end{gathered}[/tex]Therefore, the equation of the hyperbola is:
[tex]\frac{y^{2}}{36}-\frac{x^{2}}{64}=1[/tex]
