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The problem asks for the equation of the tangent line to the curve \( x^2 + 2xy^2 - y = 11 \) at the point \( (1, -2) \), using implicit differentiation.

We need to implicitly differentiate both sides of the given equation with respect to \( x \) and then solve for the slope at the point \( (1, -2) \).

### Step 1: Differentiate both sides of the equation
We start with:
\[
x^2 + 2xy^2 - y = 11
\]
Differentiate each term with respect to \( x \), using the product rule where necessary:

- \( \frac{d}{dx} (x^2) = 2x \)
- For \( 2xy^2 \), apply the product rule: 
  \[
  \frac{d}{dx}(2xy^2) = 2 \left( y^2 + x \cdot 2y \frac{dy}{dx} \right)
  \]
  This simplifies to \( 2y^2 + 4xy \frac{dy}{dx} \).
- \( \frac{d}{dx} (-y) = -\frac{dy}{dx} \).

So the differentiated equation is:
\[
2x + (2y^2 + 4xy \frac{dy}{dx}) - \frac{dy}{dx} = 0
\]

### Step 2: Substitute the point \( (1, -2) \)
Substitute \( x = 1 \) and \( y = -2 \) into the equation to find \( \frac{dy}{dx} \):

- For \( 2x \), we get \( 2(1) = 2 \),
- For \( 2y^2 \), we get \( 2(-2)^2 = 8 \),
- For \( 4xy \frac{dy}{dx} \), we get \( 4(1)(-2) \frac{dy}{dx} = -8 \frac{dy}{dx} \),
- For \( - \frac{dy}{dx} \), we get \( - \frac{dy}{dx} \).

Substituting into the differentiated equation gives:
\[
2 + 8 - 8 \frac{dy}{dx} - \frac{dy}{dx} = 0
\]
Simplify:
\[
10 - 9 \frac{dy}{dx} = 0
\]
Solve for \( \frac{dy}{dx} \):
\[
9 \frac{dy}{dx} = 10 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{10}{9}
\]

### Step 3: Equation of the tangent line
The slope of the tangent line is \( \frac{10}{9} \), and the point on the curve is \( (1, -2) \). Using the point-slope form of the equation of a line:

\[
y - (-2) = \frac{10}{9}(x - 1)
\]
Simplify:
\[
y + 2 = \frac{10}{9}x - \frac{10}{9}
\]
\[
y = \frac{10}{9}x - \frac{10}{9} - 2
\]
\[
y = \frac{10}{9}x - \frac{10}{9} - \frac{18}{9}
\]
\[
y = \frac{10}{9}x - \frac{28}{9}
\]

Thus, the equation of the tangent line is:
\[
y = \frac{10}{9}x - \frac{28}{9}
\]

Let me know if you'd like to go over any steps or explore related concepts!

Here are some follow-up questions:
1. What is implicit differentiation, and why is it used here?
2. How is the product rule applied when differentiating terms like \( 2xy^2 \)?
3. Why do we substitute the point \( (1, -2) \) after differentiating?
4. How do we simplify expressions involving fractions, like in the final equation of the tangent line?
5. What are other methods to find the equation of a tangent line, and when do they apply?

**Tip:** When differentiating implicitly, remember to apply the chain rule whenever you differentiate terms involving \( y \), since \( y \) is a function of \( x \).

By implicit differentiation, find the equation of the tangent line to the curve x^2 + 2xy^2 - y = 11 at the point (1, -2).

Learn how to use implicit differentiation to find the equation of the tangent line to the curve x^2 + 2xy^2 - y = 11 at the point (1, -2). This solution applies key concepts from calculus, including the product rule and the chain rule, to solve the problem step-by-step. Suitable for students in Grades 11-12 or early college calculus.

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