## Precalculus (10th Edition)

$\dfrac{x^2}{16}+\dfrac{y^2}{1}=1$ See graph
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: