Answer
(a)
Center: $C(0,-5)$
$V_1(0,-10)$ and $V_2(0,0)$
$F_1(0,-5-\sqrt {22})$ and $F_2(0,-5+\sqrt {22})$
(b)
Length of the major axis:
$2a=10$
Length of the minor axis:
$2b=2\sqrt 3$
(c)
Work Step by Step
Equation of an ellipse with center at $(h,k)$ (major axis is vertical):
$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$
$\frac{x^2}{3}+\frac{(y+5)^2}{25}=1$
$\frac{(x-0)^2}{(\sqrt 3)^2}+\frac{[y-(-5)]^2}{5^2}=1$
$h=0$
$k=-5$
Center: $C(0,-5)$
$a=5$
$b=\sqrt 3$
$c^2=a^2-b^2=25-3=22$
$c=\sqrt {22}$
The given equation can be obtained by shifting
$\frac{x^2}{(\sqrt 3)^2}+\frac{y^2}{5^2}=1$
downward 5 units. In this equation:
Vertices: $V(0,±a)$
$V_1(0,-5)$ and $V_2(0,5)$
Foci: $F(0,±c)$
$F_1(0,-\sqrt {22})$ and $F_2(0,\sqrt {22})$
Now, shift these points 5 units downward:
$V_1(0,-10)$ and $V_2(0,0)$
$F_1(0,-5-\sqrt {22})$ and $F_2(0,-5+\sqrt {22})$
Length of the major axis:
$2a=10$
Length of the minor axis:
$2b=2\sqrt 3$
