# Find the points on the lemniscate where the tangent is horizontal. 8(x^{2} + y^{2})^{2} = 25(x^{2 }- y^{2})

**Solution:**

Given, the equation of lemniscate is 8(x^{2 }+ y^{2})^{2} = 25(x^{2} - y^{2}) --- (1)

Differentiate with respect to x,

16(x^{2} + y^{2})(2x + 2y dy/dx) = 25(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, 16(x^{2} + y^{2})(2x + 2y(0)) = 25(2x - 2y(0))

16(x^{2} + y^{2})(2x) = 25(2x)

16(x^{2} + y^{2}) = 25

x^{2} + y^{2} = 25/16

y^{2 }= (25/16) - x^{2} --- (2)

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

8(x^{2} + (25/16) - x^{2})^{2} = 25[x^{2} - ((25/16) - x^{2})]

8(25/16)^{2 }= 25(2x^{2 }- (25/16))

On simplification,

5000/256 = 50x^{2} - 625/16

50x^{2} = (5000+10000)/256

50x^{2} = 15000/256

x^{2} = 300/256

Taking square root

x = ±(√300)/16

To find y² substitute the value of x^{2} in (2)

y^{2} = (25/16) - (300/256)

y^{2} = (400-300)/256

y^{2 }= 100/256

y^{2} = 25/64

Taking square root,

y = ±5/8

Therefore, the points on the lemniscate where the tangent is horizontal are x = ±(√300)/16 and y = ±5/8.

## Find the points on the lemniscate where the tangent is horizontal. 8(x^{2} + y^{2})^{2} = 25(x^{2} - y^{2})

**Summary:**

The points on the lemniscate 8(x^{2} + y^{2})^{2} = 25(x^{2} - y^{2}) where the tangent is horizontal are ((√300, 5/8) ((√300, -5/8) (-(√300, 5/8) and (-(√300, -5/8).