Answer
$\dfrac{(x-2)^2}{16}+\dfrac{(y-1)^2}{7}=1$
See graph
Work Step by Step
We are given the ellipse:
Foci: $(5,1),(-1,1)$
Length of major axis: $8$
Because the $y$-coordinates of the foci are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k,c$ using the foci:
$(h-c,k)=(-1,1)\Rightarrow h-c=-1,k=1$
$(h+c,k)=(5,1)\Rightarrow h+c=5$
$k=1$
$\begin{cases}
h-c=-1\\
h+c=5
\end{cases}$
$h-c+h+c=-1+5$
$2h=4$
$h=2$
$h+c=5$
$2+c=5$
$c=3$
Determine $a$ using the length of the major axis:
$2a=8$
$a=4$
Determine $b$:
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b^2=4^2-3^2$
$b^2=7$
$b=\sqrt{7}$
The equation of the ellipse is:
$\dfrac{(x-2)^2}{4^2}+\dfrac{(y-1)^2}{(\sqrt 7)^2}=1$
$\dfrac{(x-2)^2}{16}+\dfrac{(y-1)^2}{7}=1$
The center is:
$(h,k)=(2,1)$
Determine the vertices and co-vertices:
$(h-a,k)=(2-4,1)=(-2,1)$
$(h+a,k)=(2+4,1)=(6,1)$
$(h,k-b)=(2,1-\sqrt 7)$
$(h,k+b)=(2,1+\sqrt 7)$
Graph the ellipse:
