91Ó°ÊÓ

Foci at $(5,1)$and $(-1,1)$:length of the major axis is $8$.

The foci lie on the line$y=1$, so the major axis is parallel to the x-axis.

The two foci $(5,1)$and $(-1,1)$, we can find the center of the ellipse which is the midpoint of these two points.

Thus, the center of the ellipse is$(2,1)$.

The length of the major axis $2a$is given as $8$. So, we have

$$\begin{gathered} 2a=8 \\ a=4 \end{gathered}$$

The distance from the center of the ellipse $(2,1)$to the focus $(5,1)$is $c=3$

Substitute $c=3$and $a=4$in

role="math" localid="1646719438001" $$\begin{gathered} b^2=a^2-c^2 \\ {b}^2=4^2-3^2 \\ {b}^2=16-9 \\ {b}^2=7 \end{gathered}$$

The equation of ellipse is $\frac{{(x-2)}^2}{16}+\frac{{(y-1)}^2}{7}=1$

The major axis parallel to the x- axis. So the vertices of the ellipse are $a=4$units left and right of the center $(2,1)$.

Thus, the vertices are $v_1=(-2,1)$and $v_2=(6,1)$.

We use $b=\sqrt{7}$to find the two points above and below the center. The two points are $(2,1+\sqrt{7})$and $(2,1-\sqrt{7})$.