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Equation of a surface

  1. Sep 19, 2012 #1
    Question:
    a surface consists of all point P such that the distance from P to the plane y = 1 is twice the distance from P to the point Q(0,-1,0). Find the equation for this surface and identify it.

    Solution:

    the plane y = 1 is all point (x,y,z) such that y = 1.. So can i label an arbitrary point on the y plane (x,1,z)?

    I will label the point (x,1,z) Point G
    and i will assign the ordered triple (x0,y0,z0) to Point P

    If that step if valid I then came up with the following equation....

    |PG|2 = (2|GQ|)^2.

    Next I proceeded to solve for x0,y0,z0 for LHS which should give me the equation of the surface...

    Is this correct?

    I hope my explanation to my solution is understandable
     
  2. jcsd
  3. Sep 19, 2012 #2

    LCKurtz

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    Without seeing your work, I can't tell whether it is correct. It might be as long as there are no ##x_0\, y_0\, z_0## in your equation. You have the general idea. But I would start by labeling an arbitrary point on the surface by ##P=(x,y,z)##. Then the closest point on the plane ##y=1## is ##(x,1,z)##. Then do what you did. No subscripts needed.
     
  4. Sep 19, 2012 #3
    letting ##P=(x,y,z)## the distance from P to the plane y = 1 is
    |y - 1|

    Distance from P to Q(0,-1,0) is
    √[ x2 + (y+1)2 + z2]

    Now from the problem we know that

    |y - 1|2 = 4([ x2 + (y+1)2 + z2)

    y2 - 2y + 1 = 4x2 + 4y2 + 8y + 4 + 4z2
    .
    .
    -3y2 - 10y - 4x2 -4z2 = 3
    -3(y + (5/3))2 - 4x2 - 4z2 = 3 - (25/3) = -16/3

    (Divide by - 16/3)

    9/6(y + (5/3))2 + 3/4x2 + 3/4z2 = 1

    The equation satisfying the above problem is an ellipsoid with center (0,-5/3,0)
     
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