Answer
See graph
Foci: $(-\sqrt{21},0),(\sqrt{21},0)$
Work Step by Step
We are given the equation:
$\dfrac{x^2}{25}+\dfrac{y^2}{4}=1$
The standard form of the above equation is:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$:
Determine $h,k,a,b,c$
$h=0$
$k=0$
$a^2=25\Rightarrow a=\sqrt{25}=5$
$b^2=4\Rightarrow b=\sqrt 4=2$
$c^2=a^2-b^2$
$c^2=25-4$
$c^2=21$
$c=\pm\sqrt {21}$
The center of the ellipse is:
$(h,k)=(0,0)$
The endpoints of the major and minor axis are:
$(h-a,k)=(0-5,0)=(-5,0)$
$(h+a,k)=(0+5,0)=(5,0)$
$(h,k-b)=(0,0-2)=(0,-2)$
$(h,k+b)=(0,0+2)=(0,2)$
Use the center and the 4 endpoints to graph the ellipse.
Determine the foci:
$F_1(h-c,k)=(0-\sqrt{21},0)=(-\sqrt{21},0)$
$F_2(h+c,k)=(0+\sqrt{21},0)=(\sqrt{21},0)$
