Proof that the surfaces x^2+4y+z^2 = 0 , x^2+y^2+z^2-6z+7 = 0 , are in touch at p(0,-1,2)

To prove that the surfaces defined by the equations \(x^2 + 4y + z^2 = 0\) and \(x^2 + y^2 + z^2 - 6z + 7 = 0\) are tangent at the point \(P(0, -1, 2)\), we need to show that:

1. **Point \(P\) lies on both surfaces**: We will substitute the point into both surface equations and check if it satisfies them.
2. **The normal vectors to the surfaces at \(P\) are parallel**: We will calculate the gradients of the two surface functions at \(P\) and verify they are proportional.

### Step 1: Verify the Point Lies on Both Surfaces

#### Surface 1: \(x^2 + 4y + z^2 = 0\)

Substituting \(P(0, -1, 2)\) into the equation:

\[
0^2 + 4(-1) + 2^2 = 0 \implies 0 - 4 + 4 = 0
\]

This equation is satisfied, so \(P\) lies on the first surface.

#### Surface 2: \(x^2 + y^2 + z^2 - 6z + 7 = 0\)

Substituting \(P(0, -1, 2)\) into the equation:

\[
0^2 + (-1)^2 + 2^2 - 6 \cdot 2 + 7 = 0 \implies 1 + 4 - 12 + 7 = 0
\]

This equation is also satisfied, so \(P\) lies on the second surface.

### Step 2: Compute Gradients and Check if they are Parallel

The normal vector to a surface at a point is given by the gradient of its defining function. Let's compute the gradients of both surface functions.

#### Gradient of Surface 1: \(f(x, y, z) = x^2 + 4y + z^2\)

\[
\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) = (2x, 4, 2z)
\]

Evaluating at \(P(0, -1, 2)\):

\[
\nabla f(0, -1, 2) = (2 \cdot 0, 4, 2 \cdot 2) = (0, 4, 4)
\]

#### Gradient of Surface 2: \(g(x, y, z) = x^2 + y^2 + z^2 - 6z + 7\)

\[
\nabla g = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}\right) = (2x, 2y, 2z - 6)
\]

Evaluating at \(P(0, -1, 2)\):

\[
\nabla g(0, -1, 2) = (2 \cdot 0, 2 \cdot (-1), 2 \cdot 2 - 6) = (0, -2, 4 - 6) = (0, -2, -2)
\]

### Step 3: Check if the Normal Vectors are Parallel

Two vectors are parallel if one is a scalar multiple of the other. Let's check the relationship between \(\nabla f\) and \(\nabla g\):

\[
\nabla f = (0, 4, 4), \quad \nabla g = (0, -2, -2)
\]

Observe that \(\nabla g\) is \(-\frac{1}{2}\) times \(\nabla f\):

\[
(0, -2, -2) = -\frac{1}{2}(0, 4, 4)
\]

Thus, the normal vectors are parallel.

### Conclusion

Since the point \(P(0, -1, 2)\) lies on both surfaces and the gradients at that point are parallel, the surfaces are tangent at \(P\).

If you need any more clarification or have questions, feel free to ask!

### Related Questions

1. What does it mean for two surfaces to be tangent?
2. How do you find the gradient of a surface function?
3. What is the geometric interpretation of a gradient vector?
4. Can surfaces be tangent at more than one point?
5. What role does the Hessian matrix play in analyzing surface tangency?
6. How can we verify tangency for implicit surfaces?
7. How does tangent plane analysis differ for curves and surfaces?
8. How is surface curvature related to tangent planes?

### Tip

When verifying tangency of surfaces, always confirm that the gradients at the point of contact are proportional.

Learn how to prove the tangency between surfaces x^2 + 4y + z^2 = 0 and x^2 + y^2 + z^2 - 6z + 7 = 0 at point (0, -1, 2) with detailed step-by-step solutions. This problem involves verifying point existence on surfaces and checking gradients' parallelism. Suitable for advanced mathematics enthusiasts.

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