Answer
See graph
Foci: $(-5,1),(1,1)$
Work Step by Step
We are given the equation:
$\dfrac{(x+2)^2}{25}+\dfrac{(y-1)^2}{16}=1$
The standard form of the above equation is:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$:
Determine $h,k,a,b,c$:
$h=-2$
$k=1$
$a^2=25\Rightarrow a=\sqrt{25}=5$
$b^2=16\Rightarrow b=\sqrt{16}=4$
$c^2=a^2-b^2$
$c^2=25-16$
$c^2=9$
$c=\sqrt 9$
$c=3$
The center of the ellipse is:
$(h,k)=(-2,1)$
The endpoints of the major and minor axis are:
$(h-a,k)=(-2-5,1)=(-7,1)$
$(h+a,k)=(-2+5,1)=(3,1)$
$(h,k-b)=(-2,1-4)=(-2,-3)$
$(h,k+b)=(-2,1+4)=(-2,5)$
Use the center and the 4 endpoints to graph the ellipse.
Determine the foci:
$F_1(h-c,k)=(-2-3,1)=(-5,1)$
$F_2(h+c,k)=(-2+3,1)=(1,1)$
