1. The problem statement, all variables and given/known data Find the equation of the surface that is equidistant from the plane x=1, and the point (-1,0,0). 3. The attempt at a solution Okay, if I set the distance from the surface to the point, and the distance from the surface to the plane as being equal, I should have the equation. Soo..? I can tell intuitively that the surface is going to be a circular paraboloid that opens toward the point, with its vertex on the origin, but I'm not sure how to begin this one.... I tried transposing the problem down to 2 dimensions, and finding the equation of the parabola that is equidistant between a point and a line, but I keep getting lost.
Say you have the point (x, y, z). What are the expressions for the distance of that point from the plane x=1 and for the distance from that point to (-1, 0, 0)?
From point A=(x,y,z) to point (-1,0,0) is √((-1-x)^{2}+y^{2}+z^{2}) From plane to point... well I know how to find the shortest distance from the plane to the point I think. But there are an infinite amount of points on the plane, I think that's what's confusing me. But anyway, shortest distance from plane to point A=(x,y,z).. normal vector to plane is n = <1,0,0>. So if (x_{o},y_{o},z_{o}) is a point on the plane with position vector P, then (A-P) [itex]\cdot[/itex] n is the shortest distance. n is already a unit vector so I don't have to divide by the magnitude. So do I set (A-P) [itex]\cdot[/itex] n= √((-1-x)^{2}+y^{2}+z^{2})?
Hmmm. I solved and got an answer of 0 = 2x + y^2 + z^2. The computer says i'm wrong. :( (A-P) dot n = x - 1. I set that equal to the √((-1-x)^2+y^2+z^2), then square both sides and do some basic algebra. Why am I getting the wrong answer?