A surface consists of all points P such that the distance from P to the plane y=1 is twice the distance from P

Question

A surface consists of all points P such that the distance from P to the plane y=1 is twice the distance from P to the point (0,-1,0). Find an equation for this surface and identify it.

Answer 1

For a simpler illustration, determine the result on the yz coordinate plane first. Then simply rotate about the y-axis.

On the yz plane. Such a definition of the ratio of the distances is simply the reciprocal of the eccentricity.
http://en.wikipedia.org/wiki/Eccentricity_%28mathematics%29
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In such a two dimensional figure, the eccentricity is 1/2.
That means the curve is an ellipse. It has a vertex 2/3 units away from the focus and 4/3 units away from the directrix.

Extending the result to 3D now, the surface is an ellipsoid with the focus at (0,-1,0) and its corresponding vertex is at (0,-1/3,0).

Answer 2

The surface is an ellipsoid.

The point F(0, -1, 0) is one of the foci of the ellipsoid.
Let P(x, y, z) be an arbitrary point on the surface.

Distance from plane to F is:
| y - 1 |

Distance from F to P is:
√[x² + (y+1)² + z²]

We have:
2√[x² + (y+1)² + z²] = | y - 1 |

Square both sides.

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

x²/(4/3) + (y + 5/3)²/(16/9) + z²/(4/3) = 1

This is an ellipsoid with center (h, k, p) = (0, -5/3, 0).