To solve the problem, let's proceed step-by-step using implicit differentiation for the given curve:

$x^2 + 2xy^2 - y = 11$

Step 1: Implicitly Differentiate with Respect to $x$

We will differentiate both sides of the equation with respect to $x$. Remember that when differentiating terms involving $y$, we'll use the chain rule since $y$ is a function of $x$:

$\frac{d}{dx} \left( x^2 + 2xy^2 - y \right) = \frac{d}{dx} (11)$

Breaking it down term by term:

1. $\frac{d}{dx}(x^2) = 2x$
2. For the term $2xy^2$, use the product rule: $\frac{d}{dx}(2xy^2) = 2 \cdot \left( y^2 + 2xy \cdot \frac{dy}{dx} \right)$
3. $\frac{d}{dx}(-y) = -\frac{dy}{dx}$
4. The derivative of the constant $11$ is zero.

Substituting these into the equation: $2x + 2 \left( y^2 + 2xy \cdot \frac{dy}{dx} \right) - \frac{dy}{dx} = 0$

Step 2: Solve for $\frac{dy}{dx}$

Expanding the expression: $2x + 2y^2 + 4xy \cdot \frac{dy}{dx} - \frac{dy}{dx} = 0$

Rearranging to isolate $\frac{dy}{dx}$: $4xy \cdot \frac{dy}{dx} - \frac{dy}{dx} = -2x - 2y^2$ Factoring out $\frac{dy}{dx}$: $\frac{dy}{dx} (4xy - 1) = -2x - 2y^2$ $\frac{dy}{dx} = \frac{-2x - 2y^2}{4xy - 1}$

Step 3: Substitute the Point (1, -2)

Substitute $x = 1$ and $y = -2$ into the derivative: $\frac{dy}{dx} = \frac{-2(1) - 2(-2)^2}{4(1)(-2) - 1}$ $\frac{dy}{dx} = \frac{-2 - 8}{-8 - 1} = \frac{-10}{-9} = \frac{10}{9}$

So, the slope of the tangent line at the point $(1, -2)$ is $\frac{10}{9}$.

Step 4: Find the Equation of the Tangent Line

The equation of the tangent line can be written as: $y - y_1 = m(x - x_1)$ where $m = \frac{10}{9}$, and $(x_1, y_1) = (1, -2)$.

Substituting these values: $y + 2 = \frac{10}{9}(x - 1)$

Expanding this equation: $y + 2 = \frac{10}{9}x - \frac{10}{9}$ $y = \frac{10}{9}x - \frac{10}{9} - 2$ $y = \frac{10}{9}x - \frac{28}{9}$

So, the equation of the tangent line in the form $y = ax + b$ is: $y = \frac{10}{9}x - \frac{28}{9}$

Would you like more details or have any questions about this solution?

Here are five related questions you might find useful:

1. How do you use the product rule for differentiation?
2. What is the chain rule in differentiation, and when should it be used?
3. What are implicit and explicit functions in calculus?
4. How do you derive the slope of a curve at a given point?
5. What is the process for converting a linear equation into slope-intercept form?

Tip: When using implicit differentiation, always remember to apply the chain rule to any terms involving functions of $x$ and $y$.