Final answer:
The question appears to have an error in describing the surface equation, but typically, points on a surface where the tangent plane is horizontal are found by setting the partial derivatives of the function to zero and solving for the coordinates of those points.
Step-by-step explanation:
The question seems to have a typo or incorrect expression for the surface. Assuming we are dealing with a quadratic surface where we need to find the points where the tangent plane is horizontal, we would use the gradient to determine those points. The gradient vector of the function at any given point gives the direction of steepest ascent, and it is perpendicular to the level curves of that function at that point. A horizontal tangent plane would correspond to a gradient vector with no vertical component, meaning the partial derivatives with respect to all variables except z would be zero.
To find the horizontal tangent planes explicitly, we would need the correct expression for the surface. For example, if we have a surface described by a function f(x,y), the gradient of f is given by the vector (del f/del x, del f/del y). For a horizontal tangent, we need del f/del x = 0 and del f/del y = 0. Solving for x and y when these partial derivatives are zero would give us the required points. As the question doesn't provide a clear function, this procedure is a general approach that needs the actual surface equation to proceed with the exact calculations.