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To solve this problem, we need to find the points on the curve defined by

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

where the tangent lines are horizontal and vertical. We'll use **implicit differentiation** to get \(\frac{dy}{dx}\), and then examine where it is zero (horizontal tangent) and undefined (vertical tangent).

### Step 1: Differentiate Implicitly
The equation is:
\[
x^2 - 2xy + y^2 - 4x + 8 = 0
\]

Differentiating both sides with respect to \(x\):
1. For \(x^2\), the derivative is \(2x\).
2. For \(-2xy\), apply the product rule: \(-2 \left( y + x \frac{dy}{dx} \right)\).
3. For \(y^2\), the derivative is \(2y \frac{dy}{dx}\).
4. For \(-4x\), the derivative is \(-4\).
5. The constant \(8\) has a derivative of \(0\).

This gives:
\[
2x - 2 \left( y + x \frac{dy}{dx} \right) + 

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

Learn how to use implicit differentiation to find the points with horizontal and vertical tangents on the parabola defined by x^2 - 2xy + y^2 - 4x + 8 = 0. This problem is ideal for advanced math students in Grades 11-12, covering key calculus concepts.

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