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