Find the equations to the tangent planes to 7x^{2}-3y^{2}-z^{2}+21=0 which pass through the line 7x-6y+9=0, z=3

To find the equations of the tangent planes to the surface described by \( 7x^2 - 3y^2 - z^2 + 21 = 0 \) that also pass through the line given by the equations \( 7x - 6y + 9 = 0 \) and \( z = 3 \), we need to follow a systematic approach.

### Step 1: Parametrize the line

The given line is represented by two equations:

\[
7x - 6y + 9 = 0 \quad \text{and} \quad z = 3
\]

Solve for \( x \) in terms of \( y \) from the first equation:

\[
7x - 6y + 9 = 0 \implies 7x = 6y - 9 \implies x = \frac{6y - 9}{7}
\]

Thus, the parametric equations for the line are:

\[
x = \frac{6t - 9}{7}, \quad y = t, \quad z = 3
\]

where \( t \) is a parameter.

### Step 2: Implicit equation of the surface

The equation of the surface is:

\[
7x^2 - 3y^2 - z^2 + 21 = 0
\]

We can write this as:

\[
7x^2 - 3y^2 - z^2 = -21
\]

### Step 3: Gradient of the surface

The gradient of the surface gives the normal vector to the tangent plane at any point. The gradient of the function \( f(x, y, z) = 7x^2 - 3y^2 - z^2 + 21 \) is:

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

Calculating the partial derivatives:

\[
\frac{\partial f}{\partial x} = 14x, \quad \frac{\partial f}{\partial y} = -6y, \quad \frac{\partial f}{\partial z} = -2z
\]

Thus, the gradient vector at any point \( (x, y, z) \) is:

\[
\nabla f = (14x, -6y, -2z)
\]

### Step 4: Equation of the tangent plane

The equation of the tangent plane to the surface at a point \( (x_0, y_0, z_0) \) is given by:

\[
14x_0 (x - x_0) - 6y_0 (y - y_0) - 2z_0 (z - z_0) = 0
\]

We need to find the points on the surface where the tangent plane passes through the given line \( 7x - 6y + 9 = 0 \) and \( z = 3 \).

### Step 5: Check points on the surface

Since the line lies in the plane \( z = 3 \), substitute \( z = 3 \) into the equation of the surface:

\[
7x^2 - 3y^2 - 3^2 + 21 = 0
\]

Simplifying:

\[
7x^2 - 3y^2 - 9 + 21 = 0
\]

\[
7x^2 - 3y^

Explore how to find the equations of tangent planes to the surface 7x^2 - 3y^2 - z^2 + 21 = 0 that pass through the line given by 7x - 6y + 9 = 0 and z = 3. This detailed mathematical solution includes parametrization of the line, calculation of gradients, and application of the equation of a plane. Suitable for college-level students and beyond.

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