Answer
$\dfrac{(x-1)^2}{4^2}+\dfrac{(y+2)^2}{3^2}=1$
$(1-\sqrt 7,-2)$
$(1+\sqrt 7,-2)$
Work Step by Step
We are given the equation:
$9x^2+16y^2-18x+64y-71=0$
Form two squares:
$(9x^2-18x+9)+(16y^2+64y+64)-9-64-71=0$
$9(x^2-2x+1)+16(y^2+4y+4)-144=0$
$9(x-1)^2+16(y+2)^2=144$
Divide both sides by 144:
$\dfrac{9(x-1)^2}{144}+\dfrac{16(y+2)^2}{144}=1$
Simplify:
$\dfrac{(x-1)^2}{16}+\dfrac{(y+2)^2}{9}=1$
$\dfrac{(x-1)^2}{4^2}+\dfrac{(y+2)^2}{3^2}=1$
Determine $a,b,c$ from the standard equation $\dfrac{x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$:
$h=1$
$k=-2$
$a=4$
$b=3$
Determine the foci:
$c^2=a^2-b^2$
$c^2=4^2-3^2$
$c^2=7$
$c=\pm\sqrt 7$
The foci are:
$F_1(h-c,k)=(1-\sqrt 7,-2)$
$F_2(h+c,k)=(1+\sqrt 7,-2)$
