Find all the points of intersection of the surfaces whose equations are as follows:

z^2 = 2xy - 100

and

x^2 + 2y^2 - 100
---------------------
2y

To find the points of intersection between the two surfaces, we need to solve the system of equations given by the two equations:

Equation 1: z^2 = 2xy - 100
Equation 2: x^2 + 2y^2 - 100 = 2yz

To start, let's simplify Equation 2 by multiplying both sides by 2y:

2yx^2 + 4y^3 - 200y = 4yz^2

Now, let's substitute z^2 = 2xy - 100 in Equation 2:

2yx^2 + 4y^3 - 200y = 4y(2xy - 100)

Expanding and simplifying further:

2yx^2 + 4y^3 - 200y = 8xy^2 - 400y

Next, let's rearrange the terms to get all the terms on one side:

2yx^2 - 8xy^2 + 4y^3 - 200y + 400y = 0

This can be further simplified as:

2yx^2 - 8xy^2 + 4y^3 + 200y = 0

To make the equation easier to solve, let's factor out a common factor of 2y:

2y(x^2 - 4xy + 2y^2 + 100) = 0

Now, we have two separate equations:

Equation 1: z^2 = 2xy - 100
Equation 2: 2y(x^2 - 4xy + 2y^2 + 100) = 0

First, let's solve Equation 2:

2y = 0 OR (x^2 - 4xy + 2y^2 + 100) = 0

From 2y = 0, we get y = 0.

Now, let's solve the quadratic equation (x^2 - 4xy + 2y^2 + 100) = 0:

We can rewrite the equation as:
x^2 - 4xy + 2y^2 + 100 = 0

This is a quadratic equation in terms of x. We can solve it by using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 1, b = -4y, and c = 2y^2 + 100.

Now, let's substitute the values:

x = (-(-4y) ± √((-4y)^2 - 4 * 1 * (2y^2 + 100))) / (2 * 1)

Simplifying further:

x = (4y ± √(16y^2 - 8y^2 - 400)) / 2

x = (4y ± √(8y^2 - 400)) / 2

x = 2y ± √(8y^2 - 400)

Now, we have the three equations:

Equation 1: z^2 = 2xy - 100
Equation 2: y = 0
Equation 3: x = 2y ± √(8y^2 - 400)

By combining these equations, we can find the points of intersection between the two surfaces.