What Surface Is Defined by Distances to the X-Axis and YZ-Plane?

[themadhatter1]

Homework Statement

Find an equation for the surface consisting of all points p for which the distance from P to the x-axis is twice the distance from P to the yz-plane. Identify the surface.

Homework Equations

The Attempt at a Solution

$$2\sqrt{(x_p-x)^2+(y_p-0)^2+(z_p-0)^2}=\sqrt{(x_p-0)^2+(y_p-y)^2+(z_p-z)^2}$$

I square both sides simplify and move over to one side yielding:

$$3(x_p^2)+(3y_p^2+2y_py-y^2)+(3z_p^2+2zz_p-z^2)=0$$from here my natural intuition says to complete the square or factor but you can't do either. Where did I go wrong?

 

[LCKurtz]

Homework Statement

Find an equation for the surface consisting of all points p for which the distance from P to the x-axis is twice the distance from P to the yz-plane. Identify the surface.

Homework Equations

The Attempt at a Solution

$$2\sqrt{(x_p-x)^2+(y_p-0)^2+(z_p-0)^2}=\sqrt{(x_p-0)^2+(y_p-y)^2+(z_p-z)^2}$$

I square both sides simplify and move over to one side yielding:

$$3(x_p^2)+(3y_p^2+2y_py-y^2)+(3z_p^2+2zz_p-z^2)=0$$


from here my natural intuition says to complete the square or factor but you can't do either. Where did I go wrong?

Way too complicated. You don't need any p subscripts. The nearest point to (x,y,z) on the x-axis is (x,0,0) and the nearest in the yz plane is (0,y,z). Use those.

 

[themadhatter1]

Thanks.

You wind up with the cone $$4y^2+4z^2=x^2$$