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Equation of a surface

  • Thread starter Miike012
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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
 

Answers and Replies

  • #2
LCKurtz
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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
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.
 
  • #3
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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.
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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