Precalculus (10th Edition)

$\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$ See graph
We are given the ellipse: Center: $(0,0)$ Focus: $(0,-4)$ Vertex: $(0,5)$ Because the $x$-coordinates of the vertex and focus are the same, the ellipse has the equation: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ Determine $h,k$ using the center: $(h,k)=(0,0$ $h=0$ $k=0$ Determine $a$ using the vertex: $(h,k+a)=(0,5)$ $(0,0+a)=(0,5)$ $(0,a)=(0,5)$ $a=5$ Determine $c$ using the focus: $(h,k-c)=(0,-4)$ $(0,0-c)=(0,-4)$ $(0,-c)=(0,-4)$ $c=4$ Determine $b$: $a^2=b^2+c^2$ $b^2=a^2-c^2$ $b^2=5^2-4^2$ $b^2=9$ $b=\sqrt 9=3$ The equation of the ellipse is: $\dfrac{x^2}{3^2}+\dfrac{y^2}{5^2}=1$ $\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$ The center is: $(h,k)=(0,0)$ Graph the ellipse: