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

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SUMMARY

The discussion focuses on deriving the equation for a surface defined by the condition that the distance from a point P to the x-axis is twice the distance from P to the yz-plane. The correct formulation leads to the equation of a cone, specifically 4y² + 4z² = x². The initial approach involved unnecessary complexity with subscripts, which were simplified by identifying the nearest points on the x-axis and yz-plane directly.

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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)=0from here my natural intuition says to complete the square or factor but you can't do either. Where did I go wrong?
 
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themadhatter1 said:

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.
 
Thanks.

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

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