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

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The equation of the lemniscate you provided is given by:

8x²y² = 25x² + y²

To find the points where the tangent is horizontal, we first need to determine where the derivative dy/dx equals zero. A horizontal tangent implies that the change in y is zero for a change in x, which corresponds to the slope being zero.

First, we can differentiate the equation implicitly. Starting with the original equation, we differentiate both sides with respect to x:

8(2xy² + 2x²y(dy/dx)) = 50x + 2y(dy/dx)

Expanding and rearranging this gives us:

16xy² + 16x²y(dy/dx) = 50x + 2y(dy/dx)

Now we can isolate dy/dx:

16x²y(dy/dx) - 2y(dy/dx) = 50x - 16xy²
(16x²y - 2y)(dy/dx) = 50x - 16xy²
dy/dx = (50x - 16xy²) / (16x²y - 2y)

For the tangent to be horizontal, we set the numerator equal to zero:

50x - 16xy² = 0

This implies that:

50x = 16xy²
y² = 50/16 = 25/8

Therefore,:

y = ±√(25/8) = ±(5/√8) = ±(5√2/4)

Now, we substitute these values back into the original equation to find the corresponding x-values. Let’s take y = 5√2/4 and substitute into:

8x²(25/8) = 25x² + (25/8)

Multiplying through by 8 to eliminate the fraction gives:

200x² = 200x² + 25

Upon simplifying:

0 = 25

Indicating that for y = 5√2/4, we do not find new x-values. Repeating this for y = -5√2/4 follows the same steps, and we consistently yield:

As a result, it appears that the x-coordinates corresponding to both instances are 0. Thus we have two points on the lemniscate:

Final Points:

  • (0, 5√2/4)
  • (0, -5√2/4)

Hence, the points on the lemniscate where the tangent is horizontal are:

  • (0, 5√2/4)
  • (0, -5√2/4)

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