Answer
(a)
Center: $C(-1,3)$
Vertices: $V_1(-4,3)$ and $V_2(2,3)$
Foci: $F_1(-6,3)$ and $F_2(4,3)$
Asymptotes:
$y=\frac{4}{3}x+\frac{13}{3}$
and
$y=-\frac{4}{3}x+\frac{5}{3}$
(b)
Work Step by Step
Equation of a hyperbola with center at $(h,k)$ (horizontal tranverse axis):
$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$
$\frac{(x+1)^2}{9}-\frac{(y-3)^2}{16}=1$
$\frac{[x-(-1)]^2}{3^2}-\frac{(y-3)^2}{4^2}=1$
$h=-1$
$k=3$
Center: $C(-1,3)$
$a=3$
$b=4$
$c^2=a^2+b^2=3^2+4^2=25$
$c=5$
The given equation can be obtained by shifting
$\frac{x^2}{3^2}-\frac{y^2}{4^2}=1$
left 1 units and upward 3 units.
In $\frac{x^2}{3^2}-\frac{y^2}{4^2}=1$:
Vertices: $V(±a,0)$
$V_1(-3,0)$ and $V_2(3,0)$
Foci: $F(±c,0)$
$F_1(-5,0)$ and $F_2(5,0)$
Now, shift these points 1 unit to the left and 3 units upward:
Vertices: $V_1(-4,3)$ and $V_2(2,3)$
Foci: $F_1(-6,3)$ and $F_2(4,3)$
Asymptotes:
$(y-k)=±\frac{b}{a}(x-h)$
$(y-3)=±\frac{4}{3}[x-(-1)]$
$(y-3)=±\frac{4}{3}(x+1)$
$y=\frac{4}{3}x+\frac{4}{3}+3$
$y=\frac{4}{3}x+\frac{13}{3}$
and
$y=-\frac{4}{3}x-\frac{4}{3}+3$
$y=-\frac{4}{3}x+\frac{5}{3}$
