Question

In Exercises 47 and \(48,\) find an equation of the surface satisfying the conditions, and identify the surface. The set of all points equidistant from the point (0,2,0) and the plane \(y=-2\)

Short answer

The surface is a sphere with the equation of its surface as \(16=x^2 + (y-2)^2 + z^2\).

Step-by-step solution

Find the Distance from the Point to the Plane

The distance \(d\) from a point \((x_0, y_0, z_0)\) to a plane given by the equation \(Ax + By + Cz + D = 0\) can be calculated with the formula \(d = |Ax_0 + By_0 + Cz_0 + D| / \sqrt{(A^2 + B^2 + C^2)}\). Here, the equation of the plane is \(y = -2\), which can alternatively be written as \(0x + 1y + 0z + 2 = 0\). So, \(A=0, B=1, C=0, D=2\) and the point is \(0, 2, 0\) (which is \((x_0, y_0, z_0)\)). Plugging into the formula gives \(d = |0*0 + 1*2 + 0*0 + 2| / \sqrt{(0^2 + 1^2 + 0^2)} = 4\).

Find the Distance from the Point to any Other Point

The distance \(D\) from the fixed point \((x_0, y_0, z_0)\) to any other point \((x, y, z)\) is calculated using the formula \(D = \sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2}\). In this case, \(x_0=0, y_0=2, z_0=0\). So substituting them in the formula, \(D = \sqrt{(x-0)^2 + (y-2)^2 + (z-0)^2}\) which simplifies to \(D = \sqrt{x^2 + (y-2)^2 + z^2}\).

Set the Two Distances Equal

Since the surface consists of all points equidistant from the point (0,2,0) and the plane \(y=-2\), these distances can be set equal to each other to form the equation of the surface. So \(4=\sqrt{x^2 + (y-2)^2 + z^2}\).