# Finding the equation of a paraboloid

1. Sep 16, 2010

### Quisquis

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

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.

2. Sep 16, 2010

### gabbagabbahey

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

3. Sep 16, 2010

### Quisquis

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Finding equation paraboloid Date
Find the general solution of the given differential equation... Feb 16, 2018
Find f(x) which satisfies this integral function Jan 24, 2018