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.