Answer
(a)
Vertices: $V(0,±a)=V(0,±\frac{1}{2})$
Foci: $F(0,±c)=F(0,±\frac{\sqrt 5}{6})$
Eccentricity:
$e=\frac{c}{a}=\frac{\frac{\sqrt 5}{6}}{\frac{1}{2}}=\frac{\sqrt 5}{3}$
(b)
Length of the major axis:
$2a=1$
Length of the minor axis:
$2b=\frac{2}{3}$
(c)
Work Step by Step
$9x^2+4y^2=1$
$\frac{x^2}{\frac{1}{9}}+\frac{y^2}{\frac{1}{4}}=1$
$\frac{x^2}{(\frac{1}{3})^2}+\frac{y^2}{(\frac{1}{2})^2}=1$
The major axis is vertical.
Equation of an ellipse when major axis is vertical (center at the origin):
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
So:
$a=\frac{1}{2}$
$b=\frac{1}{3}$
$c^2=a^2-b^2=\frac{1}{4}-\frac{1}{9}=\frac{5}{36}$
$c=\frac{\sqrt 5}{6}$
(a)
Vertices: $V(0,±a)=V(0,±\frac{1}{2})$
Foci: $F(0,±c)=F(0,±\frac{\sqrt 5}{6})$
Eccentricity:
$e=\frac{c}{a}=\frac{\frac{\sqrt 5}{6}}{\frac{1}{2}}=\frac{\sqrt 5}{3}$
(b)
Length of the major axis:
$2a=1$
Length of the minor axis:
$2b=\frac{2}{3}$
