Find the coordinates of the point(s) at which the tangent to the curve given by 2xy−4x^2 +3y^2 = 104 is horizontal.

To find the points where the tangent to the curve \( 2xy - 4x^2 + 3y^2 = 104 \) is horizontal, we need to follow these steps:

1. **Implicit differentiation**: Differentiate the equation with respect to \(x\), treating \(y\) as a function of \(x\) (i.e., apply implicit differentiation).
2. **Set the slope of the tangent equal to zero**: The slope of the tangent is given by \(\frac{dy}{dx}\). For a horizontal tangent, \(\frac{dy}{dx} = 0\).
3. **Solve for the coordinates** of the point(s).

### Step 1: Implicit differentiation

Starting with the equation:
\[
2xy - 4x^2 + 3y^2 = 104
\]
Differentiate each term with respect to \(x\):
- For \(2xy\), apply the product rule: 
  \[
  \frac{d}{dx}(2xy) = 2 \left( x \frac{dy}{dx} + y \right)
  \]
- For \(-4x^2\), the derivative is:
  \[
  \frac{d}{dx}(-4x^2) = -8x
  \]
- For \(3y^2\), use the chain rule:
  \[
  \frac{d}{dx}(3y^2) = 6y \frac{dy}{dx}
  \]

Now, differentiating the whole equation:
\[
2 \left( x \frac{dy}{dx} + y \right) - 8x + 6y \frac{dy}{dx} = 0
\]

### Step 2: Solve for \(\frac{dy}{dx}\)

Collect terms involving \(\frac{dy}{dx}\) on one side:
\[
(2x + 6y) \frac{dy}{dx} = 8x - 2y
\]
Solving for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{8x - 2y}{2x + 6y}
\]

### Step 3: Set \(\frac{dy}{dx} = 0\)

For a horizontal tangent, \(\frac{dy}{dx} = 0\), which implies:
\[
8x - 2y = 0
\]
Solving this for \(y\):
\[
y = 4x
\]



Discover how to find the coordinates where the tangent to the curve 2xy−4x^2 +3y^2 = 104 is horizontal using implicit differentiation. Explore concepts like the slope of tangents and key calculus rules. Suitable for advanced high school calculus students.

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