Answer
See graph
Foci: $(2-4\sqrt{2},-3),(2+4\sqrt{2},-3)$
Work Step by Step
We are given the equation:
$x^2+9y^2-4x+54y+49=0$
Bring the equation to the standard form::
$(x^2-4x+4)+(9y^2+54y+81)-4-81+49=0$
$(x^2-4x+4)+9(y^2+6y+9)-36=0$
$(x-2)^2+9(y+3)^2=36$
$\dfrac{(x-2)^2}{36}+\dfrac{9(y+3)^2}{36}=1$
$\dfrac{(x-2)^2}{36}+\dfrac{(y+3)^2}{4}=1$
Determine $a,b,c$ from the standard equation $\dfrac{x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$:
$h=2$
$k=-3$
$a^2=36\Rightarrow a=\sqrt{36}=6$
$b^2=4\Rightarrow b=\sqrt 4=2$
$c^2=a^2-b^2$
$c^2=36-4$
$c^2=32$
$c=\pm\sqrt {32}$
$c=4\sqrt 2$
The center is:
$(h,k)=(2,-3)$
The endpoints of the major and minor axis are:
$(h-a,k)=(2-6,-3)=(-4,-3)$
$(h+a,k)=(2+6,-3)=(8,-3)$
$(h,k-b)=(2,-3-2)=(2,-5)$
$(h,k+b)=(2,-3+2)=(2,-1)$
The foci are:
$F_1(h-c,k)=(2-4\sqrt{2},-3)$
$F_2(h+c,k)=(2+4\sqrt{2},-3)$
