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Equation of a circular paraboloid 
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#1
Jan2512, 05:51 PM

P: 195

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. 


#2
Jan2512, 06:38 PM

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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)?



#3
Jan2512, 07:05 PM

P: 195

From point A=(x,y,z) to point (1,0,0) is
√((1x)^{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 (AP) [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 (AP) [itex]\cdot[/itex] n= √((1x)^{2}+y^{2}+z^{2})? 


#4
Jan2512, 07:12 PM

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P: 11,683

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



#5
Jan2512, 07:29 PM

P: 195

Hmmm. I solved and got an answer of 0 = 2x + y^2 + z^2.
The computer says i'm wrong. :( (AP) dot n = x  1. I set that equal to the √((1x)^2+y^2+z^2), then square both sides and do some basic algebra. Why am I getting the wrong answer? 


#6
Jan2512, 07:32 PM

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P: 11,683

Algebra error? Shouldn't it be 4x?



#7
Jan2512, 07:42 PM

P: 195

oops. i see what i did.
Thanks!! It worked. :) 


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