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Let's start by solving the problem step-by-step.

We are given the surface equation:
\[
x = y^2 + 5z^2 - 439
\]
and asked to find the equations of the tangent plane and the normal line at the point \((2, 6, -9)\).

### 1. Tangent Plane Equation:
The general formula for the equation of the tangent plane to a surface \(F(x, y, z) = 0\) at a point \((x_0, y_0, z_0)\) is:
\[
F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0
\]
Where \(F_x\), \(F_y\), and \(F_z\) are the partial derivatives of the surface equation with respect to \(x\), \(y\), and \(z\), respectively.

However, the equation is already in the form \(x = f(y, z)\), so we only need to compute the partial derivatives of the right-hand side with respect to \(y\) and \(z\).

1. Find partial derivatives:
   \[
   \frac{\partial x}{\partial y} = \frac{\partial}{\partial y}(y^2 + 5z^2 - 439) = 2y
   \]
   \[
   \frac{\partial x}{\partial z} = \frac{\partial}{\partial z}(y^2 + 5z^2 - 439) = 10z
   \]

2. Evaluate these derivatives at the point \((2, 6, -9)\):
   \[
   \frac{\partial x}{\partial y} = 2(6) = 12
   \]
   \[
   \frac{\partial x}{\partial z} = 10(-9) = -90
   \]

Now, the tangent plane has the form:
\[
(x - x_0) = \frac{\partial x}{\partial y}(y - y_0) + \frac{\partial x}{\partial z}(z - z_0)
\]
Substituting \(x_0 = 2\), \(y_0 = 6\), and \(z_0 = -9\):
\[
(x - 2) = 12(y - 6) - 90(z + 9)
\]

3. Simplify this equation to make the coefficient of \(x\) equal to 1:
\[
x - 2 = 12(y - 6) - 90(z + 9)
\]

### 2. Normal Line Equation:
The direction vector for the normal line is given by the gradient of the function:
\[
\nabla F = \left\langle 1, \frac{\partial x}{\partial y}, \frac{\partial x}{\partial z} \right\rangle = \langle 1, 12, -90 \rangle
\]
So, the parametric equation for the normal line is:
\[
\langle x, y, z \rangle = \langle 2, 6, -9 \rangle + t \langle 1, 12, -90 \rangle
\]

Thus, the equations for the normal line are:
\[
x = 2 + t
\]
\[
y = 6 + 12t
\]
\[
z = -9 - 90t
\]

Would you like more details or clarifications? Here are 5 related questions for further practice:

1. How would you find the tangent plane if the surface was implicitly given, such as \(F(x, y, z) = 0\)?
2. Can you generalize the process to find the tangent plane to any surface \(f(x, y, z) = 0\)?
3. What would happen to the normal line if we chose a different point on the surface?
4. How does the method change if the surface equation is more complex?
5. What is the geometric interpretation of the tangent plane and normal line in 3D space?

**Tip**: Always remember that the direction of the normal line is parallel to the gradient of the surface at a given point.

Find equations of the tangent plane and normal line to the surface x = y^2 + 5z^2 - 439 at the point (2, 6, -9). Tangent Plane: (make the coefficient of x equal to 1).

Find the equations of the tangent plane and normal line to the surface x = y^2 + 5z^2 - 439 at the point (2, 6, -9). This detailed solution covers the use of partial derivatives, gradients, and normal vectors to find both the tangent plane equation and the parametric equation of the normal line.

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