Use implicit differentiation to find the points where the parabola defined by
x^{2}-2xy+y^{2}+4x-8y+24 = 0
has horizontal and vertical tangent lines.

The parabola has horizontal tangent lines at the point(s) .

The parabola has vertical tangent lines at the point(s)

To find the points where the given parabola:

\[
x^2 - 2xy + y^2 + 4x - 8y + 24 = 0
\]

has horizontal and vertical tangent lines, we will use implicit differentiation.

### Step 1: Differentiate implicitly

Differentiate the equation with respect to \( x \), remembering to apply the product rule when differentiating terms involving both \( x \) and \( y \). Also, recall that \( \frac{dy}{dx} \) (denoted as \( y' \)) is the derivative of \( y \) with respect to \( x \).

\[
\frac{d}{dx}(x^2) - \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(4x) - \frac{d}{dx}(8y) + \frac{d}{dx}(24) = 0
\]

Now compute each derivative:

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

This simplifies to:

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

Group the terms involving \( \frac{dy}{dx} \):

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

Thus, we solve for \( \frac{dy}{dx} \):

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

### Step 2: Horizontal tangents

A horizontal tangent occurs when \( \frac{dy}{dx} = 0 \). So, set the numerator equal to zero:

\[
2y - 2x - 4 = 0
\]

Simplifying this:

\[
y = x + 2
\]

Substitute \( y = x + 2 \) into the original equation to find the corresponding \( x \)-coordinates:

\[
x^2 - 2x(x + 2) + (x + 2)^2 + 4x - 8(x + 2) + 24 = 0
\]

Expanding the terms:

\[
x^2 - 2x^2 - 4x + (x^2 + 4x + 4) + 4x - 8x - 16 + 24 = 0
\]

Simplifying:

\[
x^2 - 2x^2 - 4x + x^2 + 4x + 4 - 4x - 16 + 24 = 0
\]

\[
0 = 0
\]

This shows that \( y = x + 2 \) works for all \( x \). Therefore, the points where the parabola has horizontal tangents are along the line \( y = x + 2 \). To find specific points, we substitute various \( x \)-values into this line equation and verify the original equation.

### Step 3: Vertical tangents

A vertical tangent occurs when the denominator of \( \frac{dy}{dx} \) is zero. So, set the denominator equal to zero:

\[
2x - 2y - 8 = 0
\]

Simplifying this:

\[
x = y + 4
\]

Substitute \( x = y + 4 \) into the original equation to find the corresponding \( y \)-coordinates:

\[
(y + 4)^2 - 2(y + 4)y + y^2 + 4(y + 4) - 8y + 24 = 0
\]

Expanding the terms:

\[
(y^2 + 8y + 16) - 2(y^2 + 4y) + y^2 + 4y + 16 - 8y + 24 = 0
\]

Simplifying:

\[
y^2 + 8y + 16 - 2y^2 - 8y + y^2 + 4y + 16 - 8y + 24 = 0
\]

\[
0 = 0
\]

This shows that \( x = y + 4 \) works for all \( y \). Therefore, the points where the parabola has vertical tangents are along the line \( x = y + 4 \).

### Final Answer:

- The parabola has horizontal tangent lines at points along the line \( y = x + 2 \).
- The parabola has vertical tangent lines at points along the line \( x = y + 4 \).

Would you like further details or clarifications?

Here are some related questions for you to consider:
1. How does implicit differentiation help in finding tangent lines?
2. What is the geometric significance of horizontal and vertical tangent lines?
3. How can you verify a point of tangency algebraically?
4. Could the parabola have oblique tangents as well?
5. How does this problem change if additional constraints are added to the equation?

**Tip**: When working with implicit differentiation, always keep track of product rules and chain rules, especially when both \( x \) and \( y \) appear in a term!

Use implicit differentiation to find the points where the parabola defined by x^{2}-2xy+y^{2}+4x-8y+24 = 0 has horizontal and vertical tangent lines.

Learn how to use implicit differentiation to find the points where the parabola x^2 - 2xy + y^2 + 4x - 8y + 24 = 0 has horizontal and vertical tangent lines. This step-by-step solution explores key algebraic concepts, including implicit differentiation and criteria for tangent lines. Suitable for high school and college students.

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