Equation of a circular paraboloid

[ElijahRockers]

Homework Statement

Find the equation of the surface that is equidistant from the plane x=1, and the point (-1,0,0).

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.

 

[vela]

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)?

 

[ElijahRockers]

From point A=(x,y,z) to point (-1,0,0) is
√((-1-x)²+y²+z²)

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) \cdot 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) \cdot n= √((-1-x)²+y²+z²)?

 

[vela]

Yup! Since P lies in the plane x=1, you know it has the form P=(1, y', z')...

 

[ElijahRockers]

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?

 

[vela]

Algebra error? Shouldn't it be 4x?

 

[ElijahRockers]

oops. i see what i did.

Thanks! It worked. :)