1
$\begingroup$

Find an equation for the surface consisting of all points that are equidistance from the point $(-1,0,0)$ and the plane $x = 1$.

Please expalin the details. I'm lost.

  • 0
    André has explained how to go about it. To get a picture of what’s going on, consider the $2$-dimensional analogue of finding all points equidistant from the point $(-1,0)$ and the line $x=1$. If you rotate [this picture](http://en.wikipedia.org/wiki/File:Parabola_with_focus_and_directrix.svg) $90$° counterclockwise, it will show you what that looks like.2012-09-17

1 Answers 1

4

Let $(x,y,z)$ be any point. The distance from $(x,y,z)$ to $(-1,0,0)$ is, by the usual distance formula, equal to
$\sqrt{(x-(-1))^2+(y-0)^2+(z-0)^2}.\tag{$1$}$ The point on the plane $x=1$ that is nearest to $(x,y,z)$ is $(1,y,z)$. So if you apply the distance formula, the distance from $(x,y,z)$ to the plane is $\sqrt{(x-1)^2+(y-y)^2+(z-z)^2}.\tag{$2$}$
(The expression $(2)$ simplifies to $|x-1|$. One can see this more directly. For the intuition, think about the distance from $(a,b,c)$ to the "ground" $z=0$. This distance is just $|c|$.)

Set expressions $(1)$and $(2)$ equal to each other. That will give an answer, but it may be worthwhile to simplify, getting rid of silly things like $(y-y)^2$, and squaring both sides.