At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not exist, enter dne.) (x, y, z) =

Final answer:

The point on the paraboloid y = x^2 + z^2 where the tangent plane is parallel to the plane 3x + 2y + 7z = 2 is found by setting the normal vectors proportional. Solving the equations, the point is (3/4, 29/8, 7/4).

Explanation:

To find the point on the paraboloid y = x2 + z2 where the tangent plane is parallel to the plane 3x + 2y + 7z = 2, we first need to determine the normal vector of the given plane. The normal vector of the plane is defined by its coefficients, which are (3, 2, 7). For the paraboloid, we can find the normal vector at any point by taking the gradient of the function y.

The gradient of y with respect to x and z is (2x, 1, 2z). A tangent plane to the paraboloid at point (x, y, z) will have this gradient as its normal vector. To find the point where this tangent plane is parallel to the given plane, we set the gradients equal to each other up to a constant factor because parallel planes have proportional normal vectors.

Therefore, we solve the equations:

2x = 3k

1 = 2k

2z = 7k

From the second equation, k = 1/2. Substituting k into the other equations, we find x = 3/4 and z = 7/4. Now we can substitute x and z into the equation of the paraboloid to find y:

y = (3/4)2 + (7/4)2 = 9/16 + 49/16 = 58/16 = 29/8.

The point on the paraboloid where the tangent plane is parallel to the given plane is (3/4, 29/8, 7/4).

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