Answer
$\frac{x^2}{16}+\frac{(y-3)^2}{9}=1$
Work Step by Step
Step 1. Based on the graph given in the Exercise, we can identify that it is an ellipse with one of the vertex at $(4,3)$ and the origin is at one end of its minor axis.
Step 2. As the minor axis is along the y-axis, we can identify its center at $(0,3)$
Step 3. With the vertex coordinates, we get $a=4$
Step 4. As origin is at one end of the minor axis, we get $b=3$
Step 5. We can write the equation of the ellipse as $\frac{x^2}{16}+\frac{(y-3)^2}{9}=1$
