Answer
(a)
Center: $C(0,-2)$
$V_1(0,-8)$ and $V_2(0,4)$
$F_1(0,-12)$ and $F_2(0,8)$
Asymptotes:
$y=\frac{3}{4}x-2$
$y=-\frac{3}{4}x-2$
(b)
Work Step by Step
Equation of a hyperbola with center at $(h,k)$ (transverse axis is vertical):
$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$
$\frac{(y+2)^2}{36}-\frac{x^2}{64}=1$
$\frac{[y-(-2)]^2}{6^2}-\frac{(x-0)^2}{8^2}=1$
$h=0$
$k=-2$
Center: $C(0,-2)$
$a=6$
$b=8$
$c^2=a^2+b^2=6^2+8^2=36+64=100$
$c=10$
The given equation can be obtained by shifting
$\frac{y^2}{6^2}-\frac{x^2}{8^2}=1$
downward 2 units.
In $\frac{y^2}{6^2}-\frac{x^2}{8^2}=1$:
Vertices: $V(0,±a)$
$V_1(0,-6)$ and $V_2(0,6)$
Foci: $F(0,±c)$
$F_1(0,-10)$ and $F_2(0,10)$
Now, shift these points 2 units downward:
$V_1(0,-8)$ and $V_2(0,4)$
$F_1(0,-12)$ and $F_2(0,8)$
Asymptotes:
$(y−k)=±\frac{a}{b}(x−h)$
$[y-(-2)]=±\frac{6}{8}(x-0)$
$y=±\frac{3}{4}x-2$
$y=\frac{3}{4}x-2$
and
$y=-\frac{3}{4}x-2$
