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We are tasked with finding the equations of the tangent plane and the normal line to the ellipsoid:

\[
\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{9} = 3
\]

at the point \((-2, 1, -3)\).

### 1. Gradient of the ellipsoid

The general formula for the gradient of a function \( F(x, y, z) = \frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{9} - 3 = 0 \) is given by:

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

For our ellipsoid equation:

\[
\frac{\partial F}{\partial x} = \frac{2x}{4} = \frac{x}{2}, \quad \frac{\partial F}{\partial y} = \frac{2y}{9}, \quad \frac{\partial F}{\partial z} = \frac{2z}{9}
\]

Thus, the gradient at the point \((-2, 1, -3)\) is:

\[
\nabla F(-2, 1, -3) = \left( \frac{-2}{2}, \frac{2(1)}{9}, \frac{2(-3)}{9} \right) = (-1, \frac{2}{9}, -\frac{6}{9}) = (-1, \frac{2}{9}, -\frac{2}{3})
\]

### 2. Equation of the tangent plane

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

\[
\nabla F(x_0, y_0, z_0) \cdot (x - x_0, y - y_0, z - z_0) = 0
\]

Using the point \((-2, 1, -3)\) and the gradient we computed \((-1, \frac{2}{9}, -\frac{2}{3})\), we get:

\[
(-1)(x + 2) + \frac{2}{9}(y - 1) - \frac{2}{3}(z + 3) = 0
\]

Expanding this:

\[
-(x + 2) + \frac{2}{9}(y - 1) - \frac{2}{3}(z + 3) = 0
\]

Simplifying further:

\[
-x - 2 + \frac{2}{9}y - \frac{2}{9} - \frac{2}{3}z - 2 = 0
\]

Multiplying through by 9 to clear the fractions:

\[
-9x - 18 + 2y - 2 - 6z - 18 = 0
\]

\[
-9x + 2y - 6z - 38 = 0
\]

Thus, the equation of the tangent plane is:

\[
9x - 2y + 6z = -38
\]

### 3. Equation of the normal line

The normal line at \((-2, 1, -3)\) has direction given by the gradient \((-1, \frac{2}{9}, -\frac{2}{3})\). The parametric equations for the normal line are:

\[
x = -2 - t, \quad y = 1 + \frac{2}{9}t, \quad z = -3 - \frac{2}{3}t
\]

where \(t\) is the parameter.

Let me know if you'd like more details, or if you have any questions!

Here are five related questions for practice:
1. How do you find the gradient of a scalar field?
2. What is the geometric meaning of the gradient vector in relation to a surface?
3. How can you derive the equation of a tangent plane for general surfaces?
4. How do you compute parametric equations for the normal line to a surface at a given point?
5. How do you interpret the tangent plane and normal line in real-world applications?

**Tip:** When finding the gradient, be sure to compute partial derivatives carefully to avoid sign mistakes or fraction errors.

Find the equations of the tangent plane and normal line at the point (-2, 1, -3) to the ellipsoid x^2/4 + y^2/9 + z^2/9 = 3.

Learn how to find the tangent plane and normal line to the ellipsoid x^2/4 + y^2/9 + z^2/9 = 3 at the point (-2, 1, -3). This problem involves concepts from multivariable calculus, including gradients, tangent planes, and normal lines.

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