Final answer:
A tangent plane to the paraboloid y=x^2z^2 parallel to the plane 7x+4y+3z=2 does not exist because the gradient of the function defining the paraboloid does not match the normal vector of the plane at any point. Hence, the correct answer is DNE (Does Not Exist).
Step-by-step explanation:
The student asks at which point on the paraboloid y=x^2z^2 is the tangent plane parallel to the plane described by the equation 7x+4y+3z=2. To find a plane parallel to a given plane, the normal vectors of the two planes must be identical or proportional. Therefore, we need to find the gradient (normal vector) of the surface function and equate this to the normal vector of the given plane, which has components (7, 4, 3).
We start by finding the gradient of the function f(x,y,z) = y - x^2z^2, where f is a level surface of the paraboloid. The gradient of f, denoted as ((-2xz^2),1,(-2x^2z)), should be parallel to the vector (7, 4, 3). Equating the components yields the system of equations: -2xz^2 = 7k, 1 = 4k, and -2x^2z = 3k, where k is a constant scalar. Solving this system determines that the gradient is not defined for z=0, and thus the point where the plane is tangent to the paraboloid must have z != 0. By solving the system for x and z, we find that no solution exists that satisfies the equations simultaneously, as the gradient does not equal a scaled version of (7, 4, 3). Therefore, the correct answer is DNE (Does Not Exist). The nature of a plane being parallel to another plane can be a common question related to calculus and the study of surfaces. Since no real solution exists to satisfy the condition, the conclusion is that no point on the given paraboloid has a tangent plane parallel to the given plane 7x+4y+3z=2.