Answer
(a)
Center: $C(8,-6)$
Vertices: $V_1(7,-6)$ and $V_2(9,-6)$
Foci: $F_1(-\sqrt 2+8,-6)$ and $F_2(\sqrt 2+8,-6)$
Asymptotes:
$y=x-14$
and
$y=-x+2$
(b)
Work Step by Step
Equation of a hyperbola with center at $(h,k)$ (horizontal tranverse axis):
$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$
$(x-8)^2-(y+6)^2=1$
$\frac{(x-8)^2}{1^2}-\frac{[y-(-6)]^2}{1^2}=1$
$h=8$
$k=-6$
Center: $C(8,-6)$
$a=1$
$b=1$
$c^2=a^2+b^2=1^2+1^2=2$
$c=\sqrt 2$
The given equation can be obtained by shifting
$\frac{x^2}{1^2}-\frac{y^2}{1^2}=1$
right 8 units and downward 6 units.
In $\frac{x^2}{1^2}-\frac{y^2}{1^2}=1$:
Vertices: $V(±a,0)$
$V_1(-1,0)$ and $V_2(1,0)$
Foci: $F(±c,0)$
$F_1(-\sqrt 2,0)$ and $F_2(\sqrt 2,0)$
Now, shift these points 8 units to the right and 6 units downward:
Vertices: $V_1(7,-6)$ and $V_2(9,-6)$
Foci: $F_1(-\sqrt 2+8,-6)$ and $F_2(\sqrt 2+8,-6)$
Asymptotes:
$(y-k)=±\frac{b}{a}(x-h)$
$[y-(-6)]=±\frac{1}{1}(x-8)$
$(y+6)=±(x-8)$
$y=x-8-6$
$y=x-14$
and
$y=-x+8-6$
$y=-x+2$
