Answer
(a)
Vertices: $V(±4,0)$
Foci: $F(±2\sqrt {3},0)$
Eccentricity:
$e=\frac{\sqrt {3}}{2}$
(b)
Length of the major axis:
$2a=8$
Length of the minor axis:
$2b=4$
(c)
Work Step by Step
$x^2+4y^2=16$
$\frac{x^2}{16}+\frac{y^2}{4}1$
$\frac{x^2}{4^2}+\frac{y^2}{2^2}=1$
The major axis is horizontal.
Equation of an ellipse when major axis is horizontal (center at the origin):
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
So:
$a=4$
$b=2$
$c^2=a^2-b^2=4^2-2^2=16-4=12$
$c=2\sqrt {3}$
(a)
Vertices: $V(±a,0)=V(±4,0)$
Foci: $F(±c,0)=F(±2\sqrt {3},0)$
Eccentricity:
$e=\frac{c}{a}=\frac{2\sqrt {3}}{4}=\frac{\sqrt {3}}{2}$
(b)
Length of the major axis:
$2a=8$
Length of the minor axis:
$2b=4$
