Answer
See graph
Foci: $(0,-\sqrt{5}),(0,\sqrt{5})$
Work Step by Step
We are given the equation:
$9x^2+4y^2=36$
Bring the equation to the standard form:
$\dfrac{9x^2}{36}+\dfrac{4y^2}{36}=1$
$\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$
The standard form of the above equation is:
$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$:
Determine $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=9\Rightarrow a=\sqrt{9}=3$
$b^2=4\Rightarrow b=\sqrt 4=2$
$c^2=a^2-b^2$
$c^2=9-4$
$c^2=5$
$c=\sqrt 5$
The center of the ellipse is:
$(h,k)=(0,0)$
The endpoints of the major and minor axis are:
$(h,k-a)=(0,0-3)=(0,-3)$
$(h,k+a)=(0,0+3)=(0,3)$
$(h-b,k)=(0-2,0)=(-2,0)$
$(h+b,k)=(0+2,0)=(2,0)$
Use the center and the 4 endpoints to graph the ellipse.
Determine the foci:
$F_1(h,k-c)=(0,0-\sqrt{5})=(0,-\sqrt{5})$
$F_2(h,k+c)=(0,0+\sqrt{5})=(0,\sqrt{5})$
