Answer
$z=\dfrac{x^2}{4}+\dfrac{y^2}{4}+1$
Work Step by Step
We are given one point $P(0,0,2)$ and the xy -plane, that is, $(x,y,0)$.
The distance between two points can be calculated as: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$
Thus, $\sqrt{(x-0)^2+(y-0)^2+(z-2)^2}=\sqrt{(x-x)^2+(y-y)^2+(z-0)^2}$
or, $x^2+y^2+(z-2)^2=z^2$
or, $x^2+y^2-4z+4=0$
Hence, $z=\dfrac{x^2}{4}+\dfrac{y^2}{4}+1$
