Answer
(a)
Center: $C(3,-3)$
Vertices: $V_1(-1,-3)$ and $V_2(7,-3)$
Foci: $F_1(3-\sqrt {15},-3)$ and $F_2(3+\sqrt {15},-3)$
(b)
Length of the major axis:
$2a=8$
Length of the minor axis:
$2b=2$
(c)
Work Step by Step
Equation of an ellipse with center at $(h,k)$ (major axis is horizontal):
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$
$\frac{(x-3)^2}{16}+(y^2+3)=1$
$\frac{(x-3)^2}{4^2}+\frac{[y-(-3)]^2}{1^2}=1$
$h=3$
$k=-3$
Center: $C(3,-3)$
$a=4$
$b=1$
$c^2=a^2-b^2=16-1=15$
$c=\sqrt {15}$
The given equation can be obtained by shifting
$\frac{x^2}{4^2}+\frac{y^2}{1^2}=1$
right 3 units and downward 3 units. In this equation:
Vertices: $V(±a,0)$
$V_1(-4,0)$ and $V_2(4,0)$
Foci: $F(±c,0)$
$F_1(-\sqrt {15},0)$ and $F_2(\sqrt {15},0)$
Now, shift these points 3 units to the right and 3 units downward:
Vertices: $V_1(-1,-3)$ and $V_2(7,-3)$
Foci: $F_1(3-\sqrt {15},-3)$ and $F_2(3+\sqrt {15},-3)$
Length of the major axis:
$2a=8$
Length of the minor axis:
$2b=2$
