Answer
Vertices: $(0,-3),(0,3)$
Foci: $(0,-\sqrt{5}),(0,\sqrt{5})$
See graph
Work Step by Step
We are given the ellipse:
$4y^2+9x^2=36$
Put the equation in standard form:
$\dfrac{4y^2}{36}+\dfrac{9x^2}{36}=1$
$\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Determine $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=9\Rightarrow a=\sqrt{9}=3$
$b^2=4\Rightarrow b=\sqrt 4=2$
$c^2=a^2-b^2=9-4=5$
$c=\sqrt{5}$
The center is:
$(h,k)=(0,0)$
Find the vertices:
$(h,k-a)=(0,0-3)=(0,-3)$
$(h,k+a)=(0,0+3)=(0,3)$
Find the foci:
$(h,k-c)=(0,0-\sqrt{5})=(0,-\sqrt{5})$
$(h,k+c)=(0,0+\sqrt{5})=(0,\sqrt{5})$
Use the value of $b=2$ to find the two points to the left and right of the center:
$(h-b,k)=(0-2,0)=(-2,0)$
$(h+b,k)=(0+2,0)=(2,0)$
Plot the center, the vertices, the two points to the left and the right of the center, and the foci. Then, graph the ellipse:
Vertices: $(0,-3),(0,3)$
Foci: $(0,-\sqrt{5}),(0,\sqrt{5})$
See graph
