## Precalculus (10th Edition)

$\dfrac{x^2}{1}+\dfrac{y^2}{16}=1$ See graph
We are given the ellipse: Center: $(0,0)$ Vertex: $(0,4)$ $b=1$ Because the $x$-coordinates of the center and vertex 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: $(h,k+a)=(0,4)$ $(0,0+a)=(0,4)$ $(0,a)=(0,4)$ $a=4$ The equation of the ellipse is: $\dfrac{x^2}{1^2}+\dfrac{y^2}{4^2}=1$ $\dfrac{x^2}{1}+\dfrac{y^2}{16}=1$ Graph the ellipse: