Find an equation for the surface consisting of all points P = (x, y, z) whose distance to the z-axis is twice their distance from the plane z = −1.

Question

Having a lot of trouble with this one, as I can't find similar questions online or in the Stewart Calculus book.

Any help would be greatly appreciated!

Answer 1

Instead of distances, let's use distances squared. That way you can avoid square roots. So $$d_{z-axis}^2=4d_{z=-1}^2$$ The distance between a point and a line is equal to the distance between the point and the point on the line closest to the initial point. For the $z$ axis, it means the distance from $(x,y,z)$ to $(0,0,z)$. Similarly for the plane, you look at the closest point in the plane. In your case it is $(x,y,-1)$. Then $$x^2+y^2=4(z+1)^2$$ Note that if you would have a tilted axis or plane, the expression would be more complicated.

Can you see what is this surface? Call $x^2+y^2=r^2$ and take the square root.