We are tasked with finding the equation of the tangent line to the curve at the point $(2, 4)$. The given equation is:

$32(x^2 + y^2)^2 = 400xy^2$

Step 1: Implicit Differentiation

To find the slope of the tangent line, we need to differentiate both sides of the equation implicitly with respect to $x$, treating $y$ as a function of $x$ (i.e., $y = y(x)$).

Differentiate both sides:

[ \frac{d}{dx}\left[ 32(x^2 + y^2)^2 \right] = \frac{d}{dx} \left[ 400xy^2 \right] ]

1. Differentiate the left-hand side:

Using the chain rule, the derivative of $32(x^2 + y^2)^2$ is:

$32 \cdot 2(x^2 + y^2) \cdot \frac{d}{dx}(x^2 + y^2)$

Now, differentiate $x^2 + y^2$:

$\frac{d}{dx}(x^2 + y^2) = 2x + 2y \frac{dy}{dx}$

So the derivative of the left-hand side becomes:

$64(x^2 + y^2)(x + y \frac{dy}{dx})$

2. Differentiate the right-hand side:

Using the product rule, the derivative of $400xy^2$ is:

$\frac{d}{dx}(400xy^2) = 400 \left[ \frac{d}{dx}(x) \cdot y^2 + x \cdot \frac{d}{dx}(y^2) \right]$

The derivative of $x$ is 1, and the derivative of $y^2$ is $2y \frac{dy}{dx}$. So, the derivative of the right-hand side becomes:

$400(y^2 + 2xy \frac{dy}{dx})$

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

Now we have the equation:

$64(x^2 + y^2)(x + y \frac{dy}{dx}) = 400(y^2 + 2xy \frac{dy}{dx})$

Substitute $x = 2$ and $y = 4$ into this equation.

1. Compute $x^2 + y^2$:

$x^2 + y^2 = 2^2 + 4^2 = 4 + 16 = 20$

2. Substitute $x = 2$, $y = 4$:

$64(20)(2 + 4 \frac{dy}{dx}) = 400(16 + 16 \frac{dy}{dx})$

Simplify both sides:

$1280(2 + 4 \frac{dy}{dx}) = 400(16 + 16 \frac{dy}{dx})$

Expand both sides:

$2560 + 5120 \frac{dy}{dx} = 6400 + 6400 \frac{dy}{dx}$

Step 3: Solve for $\frac{dy}{dx}$

Rearrange the equation:

$5120 \frac{dy}{dx} - 6400 \frac{dy}{dx} = 6400 - 2560$

$-1280 \frac{dy}{dx} = 3840$

$\frac{dy}{dx} = \frac{3840}{-1280} = -3$

Thus, the slope of the tangent line at the point $(2, 4)$ is $-3$.

Step 4: Equation of the Tangent Line

We now use the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

Substitute $m = -3$, $x_1 = 2$, and $y_1 = 4$:

$y - 4 = -3(x - 2)$

Simplify:

$y - 4 = -3x + 6$

$y = -3x + 10$

Final Answer:

The equation of the tangent line is:

$y = -3x + 10$

Would you like more details, or do you have any questions about the solution?

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Here are 5 related questions for further exploration:

1. How do you perform implicit differentiation with more complex functions?
2. What are the steps to apply the chain rule when differentiating implicitly?
3. How does the product rule apply to functions of two variables?
4. What is the geometric meaning of a tangent line to a curve at a given point?
5. How can implicit differentiation be extended to higher-order derivatives?

Tip: Always verify your derivative calculations step by step to avoid errors when working with implicit differentiation.