Answer
(a)
Center: $C(-1,0)$
$V_1(-1,-1)$ and $V_2(-1,1)$
$F_1(-1,-\sqrt {5})$ and $F_2(-1,\sqrt {5})$
$y=\frac{1}{2}x+\frac{1}{2}$
$y=-\frac{1}{2}x-\frac{1}{2}$
(b)
Work Step by Step
Equation of a hyperbola with center at $(h,k)$ (transverse axis is vertical):
$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$
$y^2-\frac{(x+1)^2}{4}=1$
$\frac{(y-0)^2}{1^2}-\frac{[x-(-1)]^2}{2^2}=1$
$h=-1$
$k=0$
Center: $C(-1,0)$
$a=1$
$b=2$
$c^2=a^2+b^2=2^2+1^2=4+1=5$
$c=\sqrt {5}$
The given equation can be obtained by shifting
$\frac{y^2}{1^2}-\frac{x^2}{2^2}=1$
left 1 unit.
In $\frac{y^2}{1^2}-\frac{x^2}{2^2}=1$:
Vertices: $V(0,±a)$
$V_1(0,-1)$ and $V_2(0,1)$
Foci: $F(0,±c)$
$F_1(0,-\sqrt {5})$ and $F_2(0,\sqrt {5})$
Now, shift these points 1 unit downward:
$V_1(-1,-1)$ and $V_2(-1,1)$
$F_1(-1,-\sqrt {5})$ and $F_2(-1,\sqrt {5})$
Asymptotes:
$(y−k)=±\frac{a}{b}(x−h)$
$y=±\frac{1}{2}[x-(-1)]$
$y=±\frac{1}{2}(x+1)$
$y=\frac{1}{2}x+\frac{1}{2}$
and
$y=-\frac{1}{2}x-\frac{1}{2}$
