Determine an equation of the line tangent to the curve at the given point.
4 left parenthesis x squared plus y squared right parenthesis squared equals 289 xy squared4x2+y22=289xy2​; left parenthesis 4 comma 1 right parenthesis(4,1)

We are tasked with finding the equation of the tangent line to the curve \( 4(x^2 + y^2)^2 = 289xy^2 \) at the point \( (4, 1) \).

### Step 1: Differentiate the equation implicitly

We will differentiate the given equation implicitly with respect to \( x \), keeping in mind that \( y \) is a function of \( x \) (i.e., \( y = y(x) \)).

#### Original equation:
\[
4(x^2 + y^2)^2 = 289xy^2
\]

#### Differentiate both sides with respect to \( x \):

1. **Left-hand side**: Using the chain rule:
   \[
   \frac{d}{dx} \left[ 4(x^2 + y^2)^2 \right] = 8(x^2 + y^2) \cdot \left( 2x + 2y \frac{dy}{dx} \right)
   \]
   This simplifies to:
   \[
   16(x^2 + y^2)(x + y \frac{dy}{dx})
   \]

2. **Right-hand side**: Applying the product rule:
   \[
   \frac{d}{dx} \left[ 289xy^2 \right] = 289 \left( y^2 + 2xy \frac{dy}{dx} \right)
   \]

### Step 2: Substitute the point \( (4, 1) \)

Now substitute \( x = 4 \) and \( y = 1 \) into the differentiated equation to solve for \( \frac{dy}{dx} \), the slope of the tangent line at that point.

#### Simplifying each side:

- Left-hand side:
   \[
   16((4^2 + 1^2))(4 + 1 \cdot \frac{dy}{dx}) = 16(16 + 1)(4 + \frac{dy}{dx}) = 16(17)(4 + \frac{dy}{dx})
   \]
   \[
   = 272(4 + \frac{dy}{dx})
   \]

- Right-hand side:
   \[
   289 \left( 1^2 + 2(4)(1) \frac{dy}{dx} \right) = 289(1 + 8 \frac{dy}{dx})
   \]
   \[
   = 289(1 + 8 \frac{dy}{dx})
   \]

Now the equation is:
\[
272(4 + \frac{dy}{dx}) = 289(1 + 8 \frac{dy}{dx})
\]

### Step 3: Solve for \( \frac{dy}{dx} \)

Expand both sides:
\[
1088 + 272 \frac{dy}{dx} = 289 + 2312 \frac{dy}{dx}
\]
Move the terms involving \( \frac{dy}{dx} \) to one side:
\[
272 \frac{dy}{dx} - 2312 \frac{dy}{dx} = 289 - 1088
\]
\[
-2040 \frac{dy}{dx} = -799
\]
Solve for \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{799}{2040} = \frac{799}{2040} \approx 0.3917
\]

### Step 4: Equation of the tangent line

The slope of the tangent line is \( m = \frac{799}{2040} \). Using the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
where \( (x_1, y_1) = (4, 1) \), the equation becomes:
\[
y - 1 = \frac{799}{2040}(x - 4)
\]

This is the equation of the tangent line to the curve at the point \( (4, 1) \).

Would you like more details or have any questions? 

Here are 5 related questions:
1. How is implicit differentiation used to find the slope of a tangent line?
2. What is the importance of substituting the point into the derivative equation?
3. How does the chain rule apply when differentiating the left-hand side of the equation?
4. Can this method be used for any curve that is implicitly defined?
5. What is the geometric meaning of the tangent line in this context?

**Tip**: Implicit differentiation is useful when it’s difficult to solve for \( y \) explicitly in terms of \( x \).

Determine an equation of the line tangent to the curve at the given point. 4(x^2 + y^2)^2 = 289xy^2; (4,1)

Find the equation of the tangent line to the curve 4(x^2 + y^2)^2 = 289xy^2 at the point (4, 1) using implicit differentiation and advanced calculus concepts. Suitable for advanced high school or early college students.

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