## Precalculus (10th Edition)

$\dfrac{x^2}{4}+\dfrac{y^2}{13}=1$ See graph
We are given the ellipse: Foci: $(0,3),(0,3)$' $x$-intercepts: $\pm 2$ Because the foci have the same $x$-intercept, the parabola is vertical. The equation is in the form: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ Determine $h,k,b$ using the $x$-intercepts: $(h,k-b)=(0,-2)\Rightarrow h=0, k-b=-2$ $(h,k+b)=(0,2)\Rightarrow k+b=2$ $h=0$ $\begin{cases} k-b=-2\\ k+b=2 \end{cases}$ $k-b+k+b=-2+2=0$ $2k=0$ $k=0$ $k+b=2$ $0+b=2$ $b=2$ Determine $c$, using the foci: $(h,k-c)=(0,-3)$ $(0,0-c)=(0,-3)$ $(0,-c)=(0,-3)$ $c=3$ Determine $a$: $a^2=b^2+c^2$ $a^2=2^2+3^2$ $a^2=13$ $a=\sqrt{13}$ The center is: $(h,k)=(0,0)$ The equation of the ellipse is: $\dfrac{x^2}{4}+\dfrac{y^2}{13}=1$ Plot the center, the vertices, the $x$-intercepts, and the foci. Then graph the ellipse: