91Ó°ÊÓ

Vertices at $(4,3)$and $(4,9)$: focus at $(-3,0)$.

The focus and the vertex lie on the line $x=4$, so the major axis is parallel to the y-axis.

Center of the ellipse is midpoint of given vertices which is$(4,6)$.

The distance between the two vertices, that is the length of the major axis is $6$units.

The length of the major axis is also denoted as $2a$. Therefore, we have

$$\begin{gathered} 2a=6 \\ a=3 \end{gathered}$$

The distance from the center of the ellipse to the focus is $c=2$.

Substitute $a=3$and $c=2$in $b^2=a^2-c^2$we get.

$$\begin{gathered} b^2=a^2-c^2 \\ {b}^2=3^2-2^2 \\ {b}^2=9-4 \\ {b}^2=5 \\ b=\sqrt{5} \end{gathered}$$

The equation of the ellipse is$\frac{{(x-4)}^2}{5}+\frac{{(y-6)}^2}{9}=1$

Since, $c=2$and the major axis is parallel to the y-axis, the foci are $2$units up and down of the center.

Thus, the foci are $F_1=(4,4)$and$F_2=(4,8)$

Use $b=\sqrt{5}$to find the two points to the right and left of the center. The two points are $(4+\sqrt{5},6)$and $(4-\sqrt{5},6)$.

Use the vertices, foci and the points to graph the ellipse $\frac{{(x-4)}^2}{5}+\frac{{(y-6)}^2}{9}=1$