Find the points on the lemniscate where the tangent is horizontal. 2(x² + y²)² = 81(x² - y²)

Solution:

Given, the equation of lemniscate is 2(x² + y²)² = 81(x² - y²) --- (1)

Differentiate with respect to x,

4(x² + y²)(2x + 2y dy/dx) = 81(2x - 2y dy/dx)

Here, dy/dx represents slope.

We know that the slope of a horizontal line is zero.

Thus, dy/dx = 0

Now, 4(x² + y²)(2x + 2y(0)) = 81(2x - 2y(0))

4(x² + y²)(2x) = 81(2x)

4(x² + y²) = 81

x² + y² = 81/4

y² = (81/4) - x² --- (2)

Substitute the value of y² in (1)

2(x² + (81/4) - x²)² = 81[x² - ((81/4) - x²)]

2(81/4)² = 81(2x² - (81/4))

On simplification,

13122/16 = 162x² - 6561/4

162x² = (13122 + 26244)/16

162x² = 39366 /16

x² = 243/16

Taking square root

x = ±(√243)/4

To find y substitute the value of x² in (2)

y² = (81/4) - (243/16)

y² = (324 - 243)/16

y² = 81/16

Taking square root,

y = ±9/2

Therefore, the points where the tangent is horizontal are x = ±(√243)/4 and y = ±9/2.