Equation of a surface

[Miike012]

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

 

[LCKurtz]

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.

 

[Miike012]

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)