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We are tasked with finding the equation of the tangent line to the ellipse defined by the equation:

\[
5x^2 - 6xy + 5y^2 = 52
\]

at the point \((-2, -4)\) using implicit differentiation.

### Step 1: Differentiate implicitly with respect to \(x\).

We start by differentiating the given equation implicitly. For each term:

1. **\(5x^2\):**
   \[
   \frac{d}{dx}(5x^2) = 10x
   \]

2. **\(-6xy\):**
   Here, we use the product rule for differentiation:
   \[
   \frac{d}{dx}(-6xy) = -6 \left( x \frac{dy}{dx} + y \right)
   \]
   Simplifying:
   \[
   = -6 \left( \frac{dy}{dx} \cdot x + y \right)
   \]

3. **\(5y^2\):**
   Again, using the chain rule:
   \[
   \frac{d}{dx}(5y^2) = 10y \frac{dy}{dx}
   \]

4. **The constant 52:**
   Since the derivative of a constant is zero:
   \[
   \frac{d}{dx}(52) = 0
   \]

Thus, the derivative of the entire equation becomes:

\[
10x - 6 \left( x \frac{dy}{dx} + y \right) + 10y \frac{dy}{dx} = 0
\]

### Step 2: Substitute the given point \((-2, -4)\).

Substitute \(x = -2\) and \(y = -4\) into the equation to simplify:

\[
10(-2) - 6 \left( (-2) \frac{dy}{dx} + (-4) \right) + 10(-4) \frac{dy}{dx} = 0
\]
Simplifying each term:
\[
-20 - 6 \left( -2 \frac{dy}{dx} - 4 \right) + 10(-4) \frac{dy}{dx} = 0
\]
\[
-20 - 6 \left( -2 \frac{dy}{dx} - 4 \right) - 40 \frac{dy}{dx} = 0
\]
Distribute the \(-6\):
\[
-20 + 12 \frac{dy}{dx} + 24 - 40 \frac{dy}{dx} = 0
\]
Combine like terms:
\[
4 - 28 \frac{dy}{dx} = 0
\]
Solving for \(\frac{dy}{dx}\):
\[
-28 \frac{dy}{dx} = -4 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{4}{28} = \frac{1}{7}
\]

Thus, the slope of the tangent line is \(\frac{1}{7}\).

### Step 3: Use point-slope form to write the equation of the tangent line.

The point-slope form of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
Substitute \((x_1, y_1) = (-2, -4)\) and \(m = \frac{1}{7}\):
\[
y + 4 = \frac{1}{7}(x + 2)
\]
Multiply through by 7 to eliminate the fraction:
\[
7(y + 4) = x + 2
\]
\[
7y + 28 = x + 2
\]
Finally, rearrange to get the equation of the tangent line:
\[
x - 7y = 26
\]

Thus, the equation of the tangent line is:
\[
x - 7y = 26
\]

Would you like further details on any part of the solution or have any questions?

### Related Questions:
1. How does implicit differentiation differ from explicit differentiation?
2. Why do we use the product rule for the term \(-6xy\)?
3. What is the significance of the slope in the context of the tangent line?
4. Can we apply this method to any implicit equation to find tangent lines?
5. What would the equation of the normal line be at the same point?

### Tip:
Always check that the point given lies on the curve by substituting it into the original equation before proceeding with implicit differentiation.

Use implicit differentiation to find an equation of the tangent line to the ellipse defined by 5x^2 - 6xy + 5y^2 = 52 at the point (-2, -4).

Learn how to use implicit differentiation to find the equation of the tangent line to the ellipse defined by 5x^2 - 6xy + 5y^2 = 52 at the point (-2, -4). This problem involves key calculus concepts such as the product rule, chain rule, and point-slope form.

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