Answer
$\dfrac{x^2}{5}+\dfrac{y^2}{9}=1$
Work Step by Step
The standard form of an ellipse can be written as: $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1....(1); $ when $b \gt a$ has foci $(0, \pm c)$ where ; $c^2=a^2-b^2$
We are given that an ellipse with major axis is $6$ . So, $2a=6 \implies a=3$ and $a^2 =9$.and $x$-intercept at $\pm {15}$ and $y$-intercept at $\pm 4$ due to the intercepts, the centre will be at origin.
Foci are: $(0,-2)$ and $(0,2)$; This implies that $c=2$
Therefore, $c^2=a^2-b^2 \implies 2^2 =3^2-b^2 \implies b^2= 5$
Now, substitute these values into the equation (1) to obtain:
$\dfrac{x^2}{5}+\dfrac{y^2}{9}=1$
