<s>[INST] Identifying the equation of the surface containing all points that have a distance to the x-axis twice that to the yz-plane The distance $D_{1}$ of an arbitrary point P(x, y, z) to the x-axis can be obtained using the right angled triangle formed by the coordinates of the point:
$$y^2 + z^2 = D^2\\
D_{1} = \sqrt{y^2+z^2}$$
The distance $D_{2}$ from P to the yz-plane, $x = 0$, is found by:
$$D_{2} = \frac{|ax + by + cz + d|}{\sqrt{a^2 + b^2 + c^2}} = \frac{|1x + 0y + 0z + 0|}{\sqrt{1^2}} = |x|$$
Equating the two distances:
$$D_{1} = D_{2}\\
\sqrt{y^2+z^2} = 2|x|$$
The resulting equation is in the form of an elliptic cone, its axis being the x-axis:
$$y^2 + z^2 = 4x^2$$
Have I correctly identified the surface that contains all points twice the distance to the x-axis as to the yz-plane? What other methods could be used to find this surface?
 [/INST] Let
$$y^2+z^2= r^2$$
$ r= x \cdot \tan \alpha = 2 x $ is a circular cone of semi-vertical angle $\alpha= \tan^{-1}2.$
The cone is swept out by rotation of $ y= 2x \text { or } z=2 x$  about the x-axis.
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