Answer
$\dfrac{x^2}{16}+\dfrac{y^2}{1}=1$
See graph
Work Step by Step
We are given the ellipse:
Vertices: $(-4,0),(4,0)$
$y$-intercepts: $-1,1$
Because the $y$-coordinates of the vertices are the same, the ellipse has the equation:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k,b$ using the $y$-intercepts:
$(h,k-b)=(0,-1)\Rightarrow h=0,k-b=-1$
$(h,k+b)=(0,1)\Rightarrow k+b=1$
$h=0$
$\begin{cases}
k-b=-1\\
k+b=1
\end{cases}$
$k-b+k+b=-1+1$
$2k=0$
$k=0$
$k+b=1$
$0+b=1$
$b=1$
Determine $a$ using the vertices:
$(h-a,k)=(-4,0)$
$(0-a,0)=(-4,0)$
$(-a,0)=(-4,0)$
$a=4$
The equation of the ellipse is:
$\dfrac{x^2}{4^2}+\dfrac{y^2}{1^2}=1$
$\dfrac{x^2}{16}+\dfrac{y^2}{1}=1$
The center is:
$(h,k)=(0,0)$
Graph the ellipse: