Answer
$\dfrac{(x-5)^2}{21}+\dfrac{(y-2)^2}{25}=1$
Work Step by Step
The standard form of an ellipse can be written as: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^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 $10$ . So, $2a=10 \implies a=5$ and $a^2 =25$.
Foci are: $(0,-2)$ and $(0,2)$; This implies that $c=2$
Therefore, $c^2=a^2-b^2 \implies 2^2 =5^2-b^2 \implies b^2= 21$
Now, substitute these values into the equation (1) to obtain:
$\dfrac{(x-5)^2}{21}+\dfrac{(y-2)^2}{25}=1$
