Answer
$\dfrac{(x+5)^2}{16}+\dfrac{(y-1)^2}{4}=1$
Work Step by Step
We are given the equation:
$x^2+4y^2+10x-8y+13=0$
Form two squares:
$(x^2+10x+25)+(4y^2-8y+4)-25-4+13=0$
$(x^2+10x+25)+4(y^2-2y+1)-16=0$
$(x+5)^2+4(y-1)^2=16$
Divide both sides by 16:
$\dfrac{(x+5)^2}{16}+\dfrac{4(y-1)^2}{16}=1$
Simplify:
$\dfrac{(x+5)^2}{16}+\dfrac{(y-1)^2}{4}=1$
$\dfrac{(x+5)^2}{4^2}+\dfrac{(y-1)^2}{2^2}=1$
Determine $a,b,c$ from the standard equation $\dfrac{x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$:
$h=-5$
$k=1$
$a=4$
$b=2$
Determine the foci:
$c^2=a^2-b^2$
$c^2=4^2-2^2$
$c^2=12$
$c=\pm\sqrt {12}=\pm 2\sqrt 3$
The foci are:
$F_1(h-c,k)=(-5-2\sqrt 3,1)$
$F_2(h+c,k)=(-5+2\sqrt 3,1)$
