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