# Find the points on the lemniscate where the tangent is horizontal. 2(x^{2} + y^{2})^{2} = 81(x^{2} - y^{2})

**Solution:**

Given, the equation of lemniscate is 2(x^{2} + y^{2})^{2} = 81(x^{2} - y^{2}) --- (1)

Differentiate with respect to x,

4(x^{2} + y^{2})(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(x^{2} + y^{2})(2x + 2y(0)) = 81(2x - 2y(0))

4(x^{2} + y^{2})(2x) = 81(2x)

4(x^{2} + y^{2}) = 81

x^{2} + y^{2} = 81/4

y^{2} = (81/4) - x^{2} --- (2)

Substitute the value of y^{2} in (1)

2(x^{2} + (81/4) - x^{2})^{2} = 81[x^{2} - ((81/4) - x^{2})]

2(81/4)^{2} = 81(2x^{2} - (81/4))

On simplification,

13122/16 = 162x^{2} - 6561/4

162x^{2} = (13122 + 26244)/16

162x^{2} = 39366 /16

x^{2} = 243/16

Taking square root

x = ±(√243)/4

To find y substitute the value of x^{2} in (2)

y^{2} = (81/4) - (243/16)

y^{2} = (324 - 243)/16

y^{2} = 81/16

Taking square root,

y = ±9/2

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

## Find the points on the lemniscate where the tangent is horizontal. 2(x^{2} + y^{2})^{2} = 81(x^{2} - y^{2})

**Summary:**

The points on the lemniscate 2(x^{2} + y^{2})^{2} = 81(x^{2} - y^{2}) where the tangent is horizontal are ((√243, 9/2), ((√243, -9/2), (-(√243, 9/2), and (-(√243, -9/2).

visual curriculum