At what point on the ellipsoid 
x^2 + y^2 + 8z^2 = 1
 is the tangent plane parallel to the plane 
x + 2y + z = 1? 

To find the point on the ellipsoid \(x^2 + y^2 + 8z^2 = 1\) where the tangent plane is parallel to the plane \(x + 2y + z = 1\), we need to follow these steps:

1. **Identify the normal vector of the plane**: The normal vector of the plane \(x + 2y + z = 1\) is \(\mathbf{n} = (1, 2, 1)\).

2. **Find the gradient of the ellipsoid**: The gradient \(\nabla F\) of the function defining the ellipsoid \(F(x, y, z) = x^2 + y^2 + 8z^2 - 1\) gives the normal vector to the surface at any point \((x, y, z)\):
   \[
   \nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) = (2x, 2y, 16z)
   \]

3. **Set the gradients equal**: For the tangent plane to be parallel, the gradients must be proportional. Therefore, we set:
   \[
   (2x, 2y, 16z) = k(1, 2, 1)
   \]
   for some scalar \(k\). This gives us the system of equations:
   \[
   2x = k \quad (1)
   \]
   \[
   2y = 2k \quad (2)
   \]
   \[
   16z = k \quad (3)
   \]

4. **Express \(x\), \(y\), and \(z\) in terms of \(k\)**:
   From (1), \(x = \frac{k}{2}\).
   From (2), \(y = k\).
   From (3), \(z = \frac{k}{16}\).

5. **Substitute into the ellipsoid equation**:
   Substitute \(x\), \(y\), and \(z\) into the ellipsoid equation:
   \[
   \left(\frac{k}{2}\right)^2 + (k)^2 + 8\left(\frac{k}{16}\right)^2 = 1
   \]
   Simplifying this gives:
   \[
   \frac{k^2}{4} + k^2 + 8\left(\frac{k^2}{256}\right) = 1
   \]
   \[
   \frac{k^2}{4} + k^2 + \frac{k^2}{32} = 1
   \]
   Combining terms:
   \[
   \frac{8k^2}{32} + \frac{32k^2}{32} + \frac{k^2}{32} = 1
   \]
   \[
   \frac{41k^2}{32} = 1
   \]
   \[
   k^2 = \frac{32}{41} \quad \Rightarrow \quad k = \pm \sqrt{\frac{32}{41}} = \pm \frac{4\sqrt{2}}{\sqrt{41}}
   \]

6. **Calculate \(x\), \(y\), and \(z\)**:
   Substitute \(k\) back to find the points:
   \[
   x = \frac{k}{2} = \pm \frac{2\sqrt{2}}{\sqrt{41}}, \quad y = k = \pm \frac{4\sqrt{2}}{\sqrt{41}}, \quad z = \frac{k}{16} = \pm \frac{\sqrt{2}}{8\sqrt{41}}
   \]

7. **Resulting points**:
   Thus, the points on the ellipsoid where the tangent plane is parallel to the given plane are:
   \[
   \left(\frac{2\sqrt{2}}{\sqrt{41}}, \frac{4\sqrt{2}}{\sqrt{41}}, \frac{\sqrt{2}}{8\sqrt{41}}\right) \quad \text{and} \quad \left(-\frac{2\sqrt{2}}{\sqrt{41}}, -\frac{4\sqrt{2}}{\sqrt{41}}, -\frac{\sqrt{2}}{8\sqrt{41}}\right)
   \]

If you have any questions or need further details, let me know! Here are some follow-up questions you might consider:

1. How can we find the tangent plane at these points?
2. What would be the distance from the origin to these points?
3. How can we generalize this process for different planes and surfaces?
4. What are the implications of the points found in physical or geometric contexts?
5. Can we visualize this ellipsoid and the planes involved? 

**Tip**: Always verify the calculations by substituting back into the original equations.

At what point on the ellipsoid x^2 + y^2 + 8z^2 = 1 is the tangent plane parallel to the plane x + 2y + z = 1?

Explore how to determine the points on the ellipsoid x^2 + y^2 + 8z^2 = 1 where the tangent plane is parallel to the plane x + 2y + z = 1. This detailed solution utilizes gradient concepts and surface geometry, suitable for advanced high school students.

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