Finding the equation of a paraboloid

[Quisquis]

Homework Statement

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?

Homework 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.

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 \frac{1}{2},\frac{1}{2},\frac{1}{2}

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

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.

 

[gabbagabbahey]

Homework Statement

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?

Homework 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.

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 \frac{1}{2},\frac{1}{2},\frac{1}{2}

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

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.

Why not just use the point to plane distance formula and set that equal to the distance to the origin?

 

[Quisquis]

Heh... problem #1 on the homework involved just that. It didn't occur to me at all to use here though. Thanks!