Answer
$\dfrac{(x-4)^2}{9}+\dfrac{(y+2)^2}{4}=1$
Work Step by Step
We are given the equation:
$4x^2+9y^2-32x+36y+64=0$
Form two squares:
$(4x^2-32x+64)+(9y^2+36y+36)-64-36+64=0$
$4(x^2-8x+16)+9(y^2+4y+4)-36=0$
$4(x-4)^2+9(y+2)^2=36$
Divide both sides by 36:
$\dfrac{4(x-4)^2}{36}+\dfrac{9(y+2)^2}{36}=1$
Simplify:
$\dfrac{(x-4)^2}{9}+\dfrac{(y+2)^2}{4}=1$
