## Precalculus (10th Edition)

$\dfrac{x^2}{12}+\dfrac{y^2}{16}=1$ See graph
We are given the ellipse: Foci: $(0,-2),(0,2)$ Length of the major axis: $8$ Because the $x$-coordinates of the foci are the same, the ellipse has the equation: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ Determine $a$ using the length of the major axis: $2a=8$ $a=4$ Determine $h,k,c$ from the foci: $(h,k-c)=(0,-2)\Rightarrow h=0,k-c=-2$ $(h,k+c)=(0,2)\Rightarrow k+c=2$ $\begin{cases} k-c=-2\\ k+c=2 \end{cases}$ $k-c+k+c=-2+2$ $2k=0$ $k=0$ $k+c=2$ $0+c=2$ $c=2$ Determine $b$: $a^2=b^2+c^2$ $b^2=a^2-c^2$ $b^2=4^2-2^2$ $b^2=12$ $b=\sqrt{12}=2\sqrt 3$ The equation of the ellipse is: $\dfrac{x^2}{(2\sqrt 3)^2}+\dfrac{y^2}{4^2}=1$ $\dfrac{x^2}{12}+\dfrac{y^2}{16}=1$ The center is: $(h,k)=(0,0)$ Graph the ellipse: