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To solve the problem, we need to use **implicit differentiation** to find where the given parabola:

\[
x^2 - 2xy + y^2 + 6x - 10y + 25 = 0
\]

has **horizontal and vertical tangent lines**. Horizontal tangents occur when the derivative \(\frac{dy}{dx}\) is **0**, and vertical tangents occur when \(\frac{dy}{dx}\) is **undefined**.

### Step 1: Find \(\frac{dy}{dx}\) Using Implicit Differentiation
To find the derivative \(\frac{dy}{dx}\), we implicitly differentiate both sides of the equation with respect to \(x\):

\[
\frac{d}{dx} \left( x^2 - 2xy + y^2 + 6x - 10y + 25 \right) = 0
\]

Let's differentiate each term:

1. **\(x^2\)**: \(2x\)

2. **\(-2xy\)**:
   - Product rule: \(\frac{d}{dx} (uv) = u'v + uv'\)
   - Here, \(u = -2x\) and \(v = y\), so we get:
   \[
   \frac{d}{dx}(-2xy) = -2y - 2x \frac{dy}{dx}
   \]

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

4. **\(6x\)**: \(6\)

5. **\(-10y\)**: Using the chain rule,
   \[
   \frac{d}{dx}(-10y) = -10 \frac{dy}{dx}
   \]

6. **\(25\)**: \(0\)

Combining these, we have:

\[
2x - 2y - 2x \frac{dy}{dx} + 2y \frac{dy}{dx} + 6 - 10 \frac{dy}{dx} = 0
\]

### Step 2: Solve for \(\frac{dy}{dx}\)
Group the terms involving \(\frac{dy}{dx}\):

\[
-2x \frac{dy}{dx} + 2y \frac{dy}{dx} - 10 \frac{dy}{dx} = -2x + 2y - 6
\]

Factor out \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} (-2x + 2y - 10) = -2x + 2y - 6
\]

Thus:

\[
\frac{dy}{dx} = \frac{-2x + 2y - 6}{-2x + 2y - 10}
\]

### Step 3: Find Horizontal Tangents
Horizontal tangents occur when \(\frac{dy}{dx} = 0\):

\[
-2x + 2y - 6 = 0
\]

Simplify:

\[
y = x + 3
\]

### Step 4: Find Vertical Tangents
Vertical tangents occur when the denominator of \(\frac{dy}{dx}\) is \(0\):

\[
-2x + 2y - 10 = 0
\]

Simplify:

\[
y = x + 5
\]

### Step 5: Find the Points of Tangency
To find the points where the tangents are horizontal or vertical, we need to substitute these conditions back into the original equation.

#### Horizontal Tangent: \(y = x + 3\)

Substituting \(y = x + 3\) in the original equation:

\[
x^2 - 2x(x + 3) + (x + 3)^2 + 6x - 10(x + 3) + 25 = 0
\]

Expanding and simplifying:

\[
x^2 - 2x^2 - 6x + x^2 + 6x + 9 + 6x - 10x - 30 + 25 = 0
\]

\[
0 = 4
\]

This leads to no solution, indicating that there are no points with horizontal tangents.

#### Vertical Tangent: \(y = x + 5\)

Substituting \(y = x + 5\) in the original equation:

\[
x^2 - 2x(x + 5) + (x + 5)^2 + 6x - 10(x + 5) + 25 = 0
\]

Expanding and simplifying:

\[
x^2 - 2x^2 - 10x + x^2 + 10x + 25 + 6x - 10x - 50 + 25 = 0
\]

\[
0 = 0
\]

This is an identity, meaning there are infinitely many points along \(y = x + 5\) where the tangent is vertical.

### Summary of Results
- **Horizontal Tangent Lines**: There are no points with horizontal tangent lines.
- **Vertical Tangent Lines**: The parabola has vertical tangent lines at all points where \(y = x + 5\).

Do you want more details on how any of these steps were carried out, or do you have any other questions?

### Related Questions:
1. How can implicit differentiation be applied to find tangent lines for other types of curves?
2. What is the significance of horizontal and vertical tangent lines in the study of curves?
3. How can product and chain rules be systematically used in implicit differentiation?
4. Can you derive the slope function for a different implicitly defined function?
5. What conditions lead to a curve having no horizontal or vertical tangents?

**Tip**: When using implicit differentiation, always carefully apply the product and chain rules, as these are key to handling mixed \(x\) and \(y\) terms correctly.

Use implicit differentiation to find the points where the parabola defined by x^2 - 2xy + y^2 + 6x - 10y + 25 = 0 has horizontal and vertical tangent lines.

Use implicit differentiation to determine points where the parabola x^2 - 2xy + y^2 + 6x - 10y + 25 = 0 has horizontal and vertical tangents. This problem involves using product and chain rules, solving for dy/dx, and identifying tangent conditions.

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