Answer
Vertices: $(\pm 5,0)$; Major Horizontal axis length =10
Minor vertical axis length =6
$r=\sqrt {5^2-3^2}=4$; Foci:$(\pm 4,0)$
The eccentricity $e$ is given as: $e=\dfrac{4}{5}$
Work Step by Step
The general form of the equation of an ellipse can be defined as:
$\dfrac{x^2}{p^2}+\dfrac{y^2}{q^2}=1$
Here, Vertices: $(\pm p,0)$; Major Horizontal axis length =2p
Minor vertical axis length =2q
Foci:$(\pm r,0)$ and $r=\sqrt {p^2-q^2}$
The eccentricity $e$ is given as: $e=\dfrac{r}{p}$
Re-arrange the given equation in the genreal form of an ellipse as follows:
$\dfrac{x^2}{(5)^2}+\dfrac{y^2}{(3)^2}=1$
Here, we have $p=5,q=3$
Now, Vertices: $(\pm 5,0)$; Major Horizontal axis length =10
Minor vertical axis length =6
$r=\sqrt {5^2-3^2}=4$; Foci:$(\pm 4,0)$
The eccentricity $e$ is given as: $e=\dfrac{4}{5}$