Answer
$y^2+z^2=4x^2$; a circular cone.
Work Step by Step
Let us consider a point $P(x,y,z)$ and suppose $D_1$ is the distance from a point $P$ to the x-axis; that is, a distance form P to (x,0,0)
Therefore,
$D_1=\sqrt {(x-x)^2+(y-0)^2 +(z-0)^2 }=\sqrt {y^2+z^2}$
Suppose $D_2$ is the distance from a point $P$ to the yz-palne; that is, a distance form P to (0,y, z)
Therefore,
$D_1=\sqrt {(x-0)^2+(y-y)^2 +(z-z)^2 }=\sqrt {x^2}=|x|$
But $D_1=2 D_2$
so, $\sqrt {y^2+z^2}=2 |x|$
or, $y^2+z^2=4x^2$
When we set $x=k$, the traces parallel to the yz-plane that are circles with radius $2k$.
This means that $y^2+z^2=4k^2$
which is an equation of a circular cone.
