Answer
Vertices: $(0,\pm 7)$
Co-vertices: $(\pm 3,0)$
Foci: $(0,\pm 2\sqrt{10})$
Work Step by Step
It is given $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a^2=9$ (or $a=3$) and $b^2=49$ (or $b=7$).
Since $b=7$ is greater than $a=3$, the major axis is vertical and so $c=\sqrt{b^2-a^2}=\sqrt{7^2-3^2}=\sqrt{40}=2\sqrt{10}$.
Then, the ellipse has the vertices at $(0,\pm 7)$, the co-vertices at $(\pm 3,0)$, and the foci at $(0,\pm 2\sqrt{10})$.
The following is the graph of the ellipse.
