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!