Answer
$\dfrac{4x^2}{81}+\dfrac{4y^2}{17}=1$
Work Step by Step
The standard form of an ellipse can be written as: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1....(1); $and has foci $( \pm c, 0)$ where ; $c^2=a^2-b^2$.
We have the sum of the distances is $9$ units. So, $2a=9 \implies a=\dfrac{9}{2}$ and $a^2 =\dfrac{81}{4}$.
Foci are: $(-4, 0)$ and $(4,0)$; This implies that $c=4$
Therefore, $c^2=a^2-b^2 \implies 4^2 =\dfrac{81}{4}-b^2 \implies b^2= \dfrac{17}{4}$
Now, substitute these values into the equation (1) to obtain:
$\dfrac{x^2}{81/4}+\dfrac{y^2}{17/4}=1 \\ \dfrac{4x^2}{81}+\dfrac{4y^2}{17}=1$
