Answer
(a)
Center: $C(3,-1)$
Vertices: $V_1(3,-7)$ and $V_2(3,5)$
Foci: $F_1(3,-1-4\sqrt 2)$ and $F_2(3,-1+4\sqrt 2)$
(b)
Length of the major axis:
$2a=12$
Length of the minor axis:
$2b=4$
(c)
Work Step by Step
Equation of an ellipse with center at $(h,k)$ (major axis is vertical):
$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$
$9x^2-54x+y^2+2y+46=0$
$9x^2-54x+81+y^2+2y+1=-46+81+1$
$9(x-3)^2+(y+1)^2=36$
$\frac{(x-3)^2}{4}+\frac{(y+1)^2}{36}=1$
$\frac{(x-3)^2}{2^2}+\frac{[y-(-1)]^2}{6^2}=1$
$h=3$
$k=-1$
Center: $C(3,-1)$
$a=6$
$b=2$
$c^2=a^2-b^2=36-4=32$
$c=4\sqrt 2$
The given equation can be obtained by shifting
$\frac{x^2}{2^2}+\frac{y^2}{6^2}=1$
right 3 units and downward 1 unit. In this equation:
Vertices: $V(0,±a)$
$V_1(0,-6)$ and $V_2(0,6)$
Foci: $F(0,±c)$
$F_1(0,-4\sqrt 2)$ and $F_2(0,4\sqrt 2)$
Now, shift these points 3 units to the right and 1 unit downward:
Vertices: $V_1(3,-7)$ and $V_2(3,5)$
Foci: $F_1(3,-1-4\sqrt 2)$ and $F_2(3,-1+4\sqrt 2)$
Length of the major axis:
$2a=12$
Length of the minor axis:
$2b=4$
