How do you find the points on the lemniscate where the tangent is horizontal?

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The lemniscate you mentioned can be expressed in its traditional form as follows:

x2 + y2 = 1 – y2 / (x2 – 1) (a typical equation layout for a lemniscate).

However, if we consider the provided equation 2x2y2 – 81x2 + y2 = 0, we need to analyze this equation to find the points where the tangent lines are horizontal.

The first step involves determining the implicit derivatives of the equation. To find the points with horizontal tangents, we need to set the derivative of y with respect to x to zero.

  1. Start with the equation: 2x2y2 – 81x2 + y2 = 0.
  2. Differentiate both sides with respect to x:
  3. Using implicit differentiation: d/dx(2x2y2) – d/dx(81x2) + d/dx(y2) = 0.
  4. This leads to: 4xy2 + 2x2 (2y(dy/dx)) – 162x + 2y(dy/dx) = 0.
  5. We can now rearrange this for dy/dx: dy/dx = (162x – 4xy2) / (2y – 2x2).

Setting the numerator equal to zero will yield the critical points for horizontal tangents:

  • 162x – 4xy2 = 0
  • Therefore, we can find y2 = (162/4)x.
  • This simplifies to y2 = 40.5x.

Next, we’ll substitute this back into the original lemniscate equation:

  • Substituting yields: 2x2(40.5x) – 81x2 + (40.5x) = 0.
  • Simplifying, we will look for values of x that satisfy this equation, leading us to identify specific points.

After solving for x, substitute back to find the corresponding y-values using y = ±sqrt(40.5x).

This process helps discover the points (x, y) on the lemniscate where the tangents are horizontal. The exact numerical solutions can be computed for final coordinates.

In conclusion, by deriving the condition for horizontal tangents using implicit differentiation, you can pinpoint the relevant points on the lemniscate curve.

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