Find an equation for each ellipse. Graph the equation by hand.

Center at $(2,-2)$: vertex at $(7,-2)$: focus at$(4,-2)$.

Center at $(2,-2)$: Vertex at $(7,-2)$: focus at $(4,-2)$.

Since the center, focus and vertex all lies on the line $y=-2$, the major axis is parallel to the x- axis.

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

The distance from the center of the ellipse to the vertex is$a=5$.

Substitute $c=2$and $a=5$in $b^2=a^2-c^2$we get.

$$\begin{gathered} b^2=a^2-c^2 \\ {b}^2=5^2-2^2 \\ {b}^2=25-4 \\ {b}^2=21 \\ b=\pm \sqrt{21} \end{gathered}$$

The form of the equation of the ellipse with the center at $(h,k)$and the major axis parallel to the x- axis is $\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$.

we have,$h=2,k=-2,a=5$and $b=\pm \sqrt{21}$.

Therefore, the equation of the ellipse is $\frac{{(x-2)}^2}{25}+\frac{{(y+2)}^2}{21}=1$

The major axis is parallel to the x-axis, so the vertices of the ellipse are $a=5$units left and right of the center.

Thus, $V_1=(-3,-2)$and $V_2=(7,-2)$

Since, $c=2$and the major axis is parallel to the x-axis, the foci are $2$units left and right of the center.

Thus, the foci are $F_1=(0,-2)$and $F_2=(4,-2)$

We use $b=\sqrt{21}$the two points are $(2,-2+\sqrt{21})$and $(2,-2-\sqrt{21})$.

Use the vertices, foci and the points to graph the ellipse$\frac{{(x-2)}^2}{25}+\frac{{(y+2)}^2}{21}=1$.