Find an equation for the surface with all points which are equidistant of $(-1,0,0)$ and the plane $x=1$. Draw the surface.
First, this is the graph I've depicted:
Some ideas to find such equation and the corresponding graph for the surface?
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1Given a point $(x,y,z)$, can you find its distance to $(-1,0,0)$ and to the plane? And can you equate those two distances? – David Aug 12 '16 at 22:52
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The distance from $(x,y,z)$ to the plane is simply $|x-1|$, and the distance from $(x,y,z)$ to $(-1,0,0)$ is $\sqrt{(x+1)^2+y^2+z^2}$. Setting the two equal to eah other, we get
$$(x-1)^2 = (x+1)^2 + y^2 + z^2$$ $$-4x = y^2 + z^2$$ $$x = -\frac{1}{4}(y^2+z^2)$$
florence
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1Because a given point on the surface will be equidistant from the plane and $(-1,0,0)$, by definition that means the distances will be equal. – florence Aug 13 '16 at 01:06
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If $z$ is not considered standard a 2D definition of parabola can be obtained by setting equal distances as:
$$x = -\frac{y^2}{4} $$
So with z considered it becomes a 3D generalized surface of revolution as a paraboloid rotated about $x$ axis vertically up :
$$ (x,y,z) = (-2,2,2)$$
$$x = -\frac{y^2+z^2}{4} $$
$ yz $ plane is horizontal, $x$ axis is vertical as shown.
Narasimham
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so does the graph look like this https://www.dropbox.com/s/1r2hp9owr6tok6i/Taller%20No.%202%20-%201.png?dl=0 ? – InfZero Aug 17 '16 at 16:57
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