Answer
$\dfrac{(x+2)^2}{4^2}+\dfrac{(y-3)^2}{8^2}=1$
$(-2,3-4\sqrt 3)$
$(-2,3+4\sqrt 3)$
Work Step by Step
We are given the equation:
$4x^2+y^2+16x-6y-39=0$
Form two squares:
$(4x^2+16x+16)+(y^2-6y+9)-16-9-39=0$
$4(x^2+4x+4)+(y^2-6y+9)-64=0$
$4(x+2)^2+(y-3)^2=64$
Divide both sides by 64:
$\dfrac{4(x+2)^2}{64}+\dfrac{(y-3)^2}{64}=1$
Simplify:
$\dfrac{(x+2)^2}{16}+\dfrac{(y-3)^2}{64}=1$
$\dfrac{(x+2)^2}{4^2}+\dfrac{(y-3)^2}{8^2}=1$
Determine $a,b,c$ from the standard equation $\dfrac{x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$:
$h=-2$
$k=3$
$a=8$
$b=4$
Determine the foci:
$c^2=a^2-b^2$
$c^2=8^2-4^2$
$c^2=48$
$c=\pm\sqrt {48}=\pm 4\sqrt 3$
The foci are:
$F_1(h,k-c)=(-2,3-4\sqrt 3)$
$F_2(h,k+c)=(-2,3+4\sqrt 3)$