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We are given the equation of a curve:

\[
x^2 - 2xy + y^2 + 4x - 8y + 16 = 0
\]

We are tasked with finding the points where this curve has horizontal and vertical tangent lines using **implicit differentiation**. 

### Steps:

1. **Implicitly differentiate** both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\) (i.e., apply the chain rule to terms involving \(y\)).
2. **Set the slope of the tangent line \( \frac{dy}{dx} \) equal to zero** for horizontal tangents (since a horizontal line has zero slope).
3. **Find where \( \frac{dy}{dx} \) is undefined** for vertical tangents (as the slope becomes infinite for vertical lines).

Let’s begin with implicit differentiation:

\[
\frac{d}{dx}\left(x^2 - 2xy + y^2 + 4x - 8y + 16\right) = \frac{d}{dx}(0)
\]

We'll differentiate each term one by one.

Let me solve this.After performing implicit differentiation, the result is:

\[
\left( -2x + 2y - 8 \right) \frac{dy}{dx} + 2x - 2y + 4 = 0
\]

Now, we can solve for \( \frac{dy}{dx} \), the slope of the tangent line:

\[
\frac{dy}{dx} = \frac{- (2x - 2y + 4)}{-2x + 2y - 8}
\]

Next, let's use this expression for the following conditions:

1. **Horizontal tangents**: Set \( \frac{dy}{dx} = 0 \) and solve for the points.
2. **Vertical tangents**: Find where the denominator \( -2x + 2y - 8 = 0 \), which makes \( \frac{dy}{dx} \) undefined.

I'll now solve for both cases.The solutions for the tangents are:

1. **Horizontal tangents**: The condition \( \frac{dy}{dx} = 0 \) gives the equation \( y = x + 2 \).
2. **Vertical tangents**: The condition where \( \frac{dy}{dx} \) is undefined (denominator equals 0) gives the equation \( y = x + 4 \).

Next, we'll substitute these equations back into the original curve equation to find the specific points where horizontal and vertical tangents occur. Let's compute that now.The specific points where the parabola has horizontal and vertical tangents are:

- **Horizontal tangent**: The point \((1, 3)\).
- **Vertical tangent**: The point \((0, 4)\).

Would you like more details on any of the steps, or have any other questions?

### Related Questions:
1. How do you interpret the slope of a curve geometrically?
2. What is implicit differentiation, and how is it different from regular differentiation?
3. Can a curve have both a horizontal and vertical tangent at the same point?
4. How would you approach finding tangents for other types of conics (circles, ellipses, etc.)?
5. What are the applications of finding horizontal and vertical tangents in real-world problems?

### Tip:
When solving implicit differentiation problems, always double-check your steps when applying the chain rule, especially when multiple terms involve both \(x\) and \(y\).

Use implicit differentiation to find the points where the parabola defined by x^2 - 2xy + y^2 + 4x - 8y + 16 = 0 has horizontal and vertical tangent lines.

Learn how to find points where the curve defined by x^2 - 2xy + y^2 + 4x - 8y + 16 has horizontal and vertical tangent lines using implicit differentiation. This problem is suitable for Grades 11-12 and calculus students, covering key concepts such as implicit differentiation and tangent lines.

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