Surface with specifications

  • #1

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



[tex]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}[/tex]

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

[tex]3(x_p^2)+(3y_p^2+2y_py-y^2)+(3z_p^2+2zz_p-z^2)=0[/tex]


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

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,559
770

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



[tex]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}[/tex]

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

[tex]3(x_p^2)+(3y_p^2+2y_py-y^2)+(3z_p^2+2zz_p-z^2)=0[/tex]


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.
 
  • #3
Thanks.

You wind up with the cone [tex]4y^2+4z^2=x^2[/tex]
 

Related Threads on Surface with specifications

  • Last Post
Replies
4
Views
984
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
926
  • Last Post
Replies
5
Views
984
Replies
2
Views
5K
  • Last Post
Replies
8
Views
1K
Replies
2
Views
9K
  • Last Post
Replies
10
Views
968
Top