Answer
$1=\dfrac{x^2}{4/3}+\dfrac{(y+5/3)^2}{16/9}+\dfrac{z^2}
{4/3}$
It is an ellipsoid centered at $ ( 0, -\frac{5}{3}, 0)$
Work Step by Step
Use the distance formula from P to plane $y=1$ is twice distance from $P$ to point $( 0,-1,0)$ .
$\sqrt {(x-x)^2+(y-1)^2+(z-z)^2}=2\sqrt {\sqrt {(x-0)^2+(y-(-1))^2+(z-0)^2}}$
$(y-1)^2=4 (x^2+4y^2+8y+4+4z^2$
$y^2-2y+1=4 (x^2+4y^2+8y+4+4z^2$
$-3=4x^2+3(y^2+10/3y+(10/6)^2)+4z^2$
$1=\dfrac{x^2}{4/3}+\dfrac{(y+5/3)^2}{16/9}+\dfrac{z^2}
{4/3}$
It is an ellipsoid centered at $ ( 0, -\frac{5}{3}, 0)$