Answer
$\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$
Work Step by Step
The general form of the equation of an ellipse can be defined as:
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ when $a \gt b$
Here, we have Major axis length:'$2a=12 \implies a=6$
or, $a^2=(6)^2 =36$
The eccentricity $e$ is given as: $e=\dfrac{c}{a}=\dfrac{\sqrt 5}{3}$
so, $\dfrac{c}{6}=\dfrac{\sqrt 5}{3} \implies c=2 \sqrt 5$
$c=\sqrt {a^2-b^2} \implies 2 \sqrt 5=\sqrt {36-b^2} $
or, $b^2=16$
Thus, the general form of an ellipse as follows:
$\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$
