Answer
$\dfrac{x^2}{3}+\dfrac{y^2}{4}=1$
See graph
Work Step by Step
We are given the ellipse:
Center: $(0,0)$
Focus: $(0,1)$
Vertex: $(0,-2)$
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:
$(h,k-a)=(0,-2)$
$(0,0-a)=(0,-2)$
$(0,-a)=(0,-2)$
$a=2$
Determine $c$ using the focus:
$(h,k+c)=(0,1)$
$(0,0+c)=(0,1)$
$(0,c)=(0,1)$
$c=1$
Determine $b$:
$a^2=b^2+c^2$
$b^2=a^2-c^2$
$b^2=2^2-1^2$
$b^2=3$
$b=\sqrt 3$
The equation of the ellipse is:
$\dfrac{x^2}{(\sqrt 3)^2}+\dfrac{y^2}{2^2}=1$
$\dfrac{x^2}{3}+\dfrac{y^2}{4}=1$
The center is:
$(h,k)=(0,0)$
Graph the ellipse:
