Answer
Vertices: $(\pm 5,0)$
Major Horizontal axis length $=10$
Minor vertical axis length $=6$
$c=4$
Foci:$F(\pm 4,0)$
$e=\dfrac{4}{5}$
Work Step by Step
Re-write the given equation in the general form of an ellipse as follows:
$\dfrac{x^2}{(5)^2}+\dfrac{y^2}{(3)^2}=1$
or, $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
Here, we get $a=5,b=3$
Thus, we have
Vertices: $(\pm 5,0)$
Major Horizontal axis length $=10$
Minor vertical axis length $=6$
$c=\sqrt {a^2-b^2}=\sqrt {5^2-3^2}=4$
Foci:$F(\pm 4,0)$
The eccentricity $e$ is given by $e=\dfrac{c}{a}=\dfrac{4}{5}$
