Answer
Vertices: $(0,\pm 5)$; Major vertical axis length =10
Minor horizontal axis length =8
$r=\sqrt {5^2-4^2}=3$; Foci:$(0,\pm 3)$
The eccentricity $e$ is given as: $e=\dfrac{3}{5}$
Work Step by Step
The general form of the equation of an vertical ellipse can be defined as:
$\dfrac{x^2}{q^2}+\dfrac{y^2}{p^2}=1$
Here, Vertices: $(0,\pm q)$; Major vertical axis length =2p
Minor horizontal axis length =2q
Foci:$(0, \pm r)$ 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:
$\dfrac{x^2}{(4)^2}+\dfrac{y^2}{(5)^2}=1$
Here, we have $p=5,q=4$
Now, Vertices: $(0,\pm 5)$; Major vertical axis length =10
Minor horizontal axis length =8
$r=\sqrt {5^2-4^2}=3$; Foci:$(0,\pm 3)$
The eccentricity $e$ is given as: $e=\dfrac{3}{5}$
