Answer
$\dfrac{x^2}{4/3}+\dfrac{(y+5/3)^2}{16/9}+\dfrac{z^2}{4/3}=1$
We have an ellipsoid centered at $(0, \dfrac{-5}{3}, 0)$
Work Step by Step
We use the distance formula:
$(y-1)=2 \sqrt {x^2+(y+1)^2+z^2}$
$\implies y^2-2y+1=4x^2+4y^2+8y+4+4z^2$
$\implies -3=4x^2 +3(y^2+\dfrac{10 y}{3})+4z^2)$
$\implies -3+3(\dfrac{10}{6})^2=4x^2 +3(y^2+\dfrac{10 y}{3}(\dfrac{10}{6})^2))+4z^2)$
$\implies \dfrac{16}{3} =4x^2 +3(y+\dfrac{5}{3})^2+4z^2$
$\implies \dfrac{x^2}{4/3}+\dfrac{(y+5/3)^2}{16/9}+\dfrac{z^2}{4/3}=1$
We see that we have an ellipsoid centered at $(0, \dfrac{-5}{3}, 0)$
