To find the equation of the tangent line to the curve given by the implicit equation $x^2 + 6xy + 12y^2 = 28$ at the point $(2, 1)$, we can proceed as follows:

Step 1: Differentiate implicitly with respect to $x$.

Given: $x^2 + 6xy + 12y^2 = 28$ we'll take the derivative of each term with respect to $x$, using implicit differentiation.

1. For $x^2$: $\frac{d}{dx}(x^2) = 2x$

2. For $6xy$: $\frac{d}{dx}(6xy) = 6 \left( \frac{dx}{dx} \cdot y + x \cdot \frac{dy}{dx} \right) = 6(y + x \cdot y')$

3. For $12y^2$: $\frac{d}{dx}(12y^2) = 12 \cdot 2y \cdot \frac{dy}{dx} = 24y y'$

Putting it all together, we get: $2x + 6(y + x y') + 24y y' = 0$

Step 2: Substitute $x = 2$ and $y = 1$ into the differentiated equation and solve for $y'$.

Substitute $x = 2$ and $y = 1$ into the equation: $2(2) + 6(1 + 2y') + 24(1)y' = 0$ which simplifies to: $4 + 6 + 12y' + 24y' = 0$ $10 + 36y' = 0$ $36y' = -10$ $y' = -\frac{5}{18}$

So, the slope of the tangent line at the point $(2, 1)$ is $-\frac{5}{18}$.

Step 3: Use the point-slope form to write the equation of the tangent line.

The point-slope form of a line is: $y - y_1 = m(x - x_1)$ Substitute $m = -\frac{5}{18}$, $x_1 = 2$, and $y_1 = 1$: $y - 1 = -\frac{5}{18}(x - 2)$ This is the equation of the tangent line.

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Follow-Up Questions

1. How would the solution change if the point were different?
2. What are other applications of implicit differentiation?
3. How does implicit differentiation differ from explicit differentiation?
4. Can this method be used for other types of curves, like hyperbolas?
5. What is the geometric interpretation of the tangent line slope?

Tip

Always check if the point lies on the curve before calculating the tangent to ensure it's a valid point on the function.