we have the equation
[tex]\frac{(y-4)^2}{4}-\frac{(x+8)^2}{9}=1[/tex]
Part 1
Find out the center
The center is the ordered pair (-8,4)
Part 2
we have that
the transverse axis lies on the y-axis
a^2=4 ----------> a=2
b^2=9 --------> b=3
The coordinates of the vertices are
(-8,4+2) --------> (-8,6)
(-8,4-2) -------> (-8,2)
The vertices are (-8,6) and (-8,2)
Part 3
Find out the coordinates of Foci
Remember that
c^2=a^2+^2
c^2=4+9
c^2=13
c=√13
The coordinates of Foci are
(-8, 4+√13) and (-8,4-√13)
Part 4
Find out the equation of the asymptotes
The equation of te asymptotes is given bty
[tex]y-k=\pm\frac{a}{b}(x-h)[/tex]
where
h=-8
k=4
a=2
b=3
substitute
[tex]y-4=\operatorname{\pm}\frac{2}{3}(x+8)[/tex]
therefore
the equations are
[tex]\begin{gathered} y=\frac{2}{3}(x+8)+4=\frac{2}{3\text{ }}x+\frac{16}{3}+4=\frac{2}{3\text{ }}x+\frac{28}{3} \\ y=-\frac{2}{3}(x+8)+4=-\frac{2}{3}x-\frac{16}{3}+4=-\frac{2}{3}x-\frac{4}{3} \end{gathered}[/tex]
The asymptotes are
[tex]\begin{gathered} y=\frac{2}{3\text{ }}x+\frac{28}{3} \\ \\ y=-\frac{2}{3}x-\frac{4}{3} \end{gathered}[/tex]
