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Homework Help: Finding the equation of a paraboloid

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Find an equation of the form Ax2+By2+Cz2+Dxy+Exz+Fyz+Gx+Hy+Jz+K=0 Satisfied by the set of all points in space, (x,y,z), whose distance to the origin is equal to their distance to the plane x+y+z=3. Based on what you know about parabolas, what does this collection of points look like?

    2. Relevant equations

    x+y+z=3

    Equation of a paraboloid: z/c=x2/a2+y2/b2

    a(x-x0)+b(y-y0)+c(z-z0)=0 The coefficients (a,b,c) is the normal vector to the plane.


    3. The attempt at a solution

    I started by finding a point that lies on the plane. The point (1,1,1) satisfies the given equation: x+y+z=3.

    Given that point, I can work back to the normal vector:

    (x-1)+(y-1)+(z-1)=0

    The normal vector is the coefficients of this eqn, so the normal vector is <1,1,1>.

    The focus is given as (0,0,0), so the vertex of the paraboloid should be [tex]\frac{1}{2},\frac{1}{2},\frac{1}{2}[/tex]

    The distance from the origin (and thus the plane) to the vertex of the paraboloid is 1/2[tex]\left\|N\right\|[/tex]=[tex]\frac{\sqrt{3}}{2}[/tex]

    That's as far as I've gotten... I really have no idea how to go from here. I think I've got all the info to put it together, I just don't know how.

    BTW, sorry about any formatting snafus, Google Chrome doesn't play well with latex at all.
     
  2. jcsd
  3. Sep 16, 2010 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Why not just use the point to plane distance formula and set that equal to the distance to the origin?
     
  4. Sep 16, 2010 #3
    Heh... problem #1 on the homework involved just that. It didn't occur to me at all to use here though. Thanks!
     
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