Answer
(a)
Center: $C(-3,1)$
$V_1(-3,-4)$ and $V_2(-3,6)$
$F_1(-3,1-\sqrt {26})$ and $F_2(-3,1+\sqrt {26})$
$y=5x+16$
$y=-5x-14$
(b)
Work Step by Step
Equation of a hyperbola with center at $(h,k)$ (transverse axis is vertical):
$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$
$\frac{(y-1)^2}{25}-(x+3)^2=1$
$\frac{(y-1)^2}{5^2}-\frac{[x-(-3)]^2}{1^2}=1$
$h=-3$
$k=1$
Center: $C(-3,1)$
$a=5$
$b=1$
$c^2=a^2+b^2=5^2+1^2=25+1=26$
$c=\sqrt {26}$
The given equation can be obtained by shifting
$\frac{y^2}{5^2}-\frac{x^2}{1^2}=1$
left 3 units and upward 1 unit.
In $\frac{y^2}{5^2}-\frac{x^2}{1^2}=1$:
Vertices: $V(0,±a)$
$V_1(0,-5)$ and $V_2(0,5)$
Foci: $F(0,±c)$
$F_1(0,-\sqrt {26})$ and $F_2(0,\sqrt {26})$
Now, shift these points 3 units to the left and 1 unit upward:
$V_1(-3,-4)$ and $V_2(-3,6)$
$F_1(-3,1-\sqrt {26})$ and $F_2(-3,1+\sqrt {26})$
Asymptotes:
$(y−k)=±\frac{a}{b}(x−h)$
$y-1=±\frac{5}{1}[x-(-3)]$
$y=±5(x+3)+1$
$y=5x+16$
and
$y=-5x-14$
