Answer
(a)
Center: $C(-1,4)$
$V_1(-1,-2)$ and $V_2(1,10)$
$F_1(-1,4-2\sqrt {10})$ and $F_2(-1,4+2\sqrt {10})$
$y=3x+7$
$y=-3x+1$
(b)
Work Step by Step
Equation of a hyperbola with center at $(h,k)$ (transverse axis is vertical):
$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$
$36x^2+72x-4y^2+32y+116=0$
$36x^2+72x+36-4y^2+32y-64=-116+36-64$
$36(x+1)^2-4(y-4)^2=-144$
$\frac{(y-4)^2}{36}-\frac{(x+1)^2}{4}=1$
$\frac{(y-4)^2}{6^2}-\frac{[x-(-1)]^2}{2^2}=1$
$h=-1$
$k=4$
Center: $C(-1,4)$
$a=6$
$b=2$
$c^2=a^2+b^2=6^2+2^2=36+4=40$
$c=2\sqrt {10}$
The given equation can be obtained by shifting
$\frac{y^2}{6^2}-\frac{x^2}{2^2}=1$
left 1 unit and upward 4 units.
In $\frac{y^2}{6^2}-\frac{x^2}{2^2}=1$:
Vertices: $V(0,±a)$
$V_1(0,-6)$ and $V_2(0,6)$
Foci: $F(0,±c)$
$F_1(0,-2\sqrt {10})$ and $F_2(0,2\sqrt {10})$
Now, shift these points 1 unit to the left and 8 units upward:
$V_1(-1,-2)$ and $V_2(1,10)$
$F_1(-1,4-2\sqrt {10})$ and $F_2(-1,4+2\sqrt {10})$
Asymptotes:
$(y−k)=±\frac{a}{b}(x−h)$
$y-4=±\frac{6}{2}[x-(-1)]$
$y-4=±3(x+1)$
$y=3x+7$
and
$y=-3x+1$
