Find the points on the lemniscate where the tangent is horizontal. 8(x2 + y2)2 = 25(x2 - y2)
Solution:
Given, the equation of lemniscate is 8(x2 + y2)2 = 25(x2 - y2) --- (1)
Differentiate with respect to x,
16(x2 + y2)(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(x2 + y2)(2x + 2y(0)) = 25(2x - 2y(0))
16(x2 + y2)(2x) = 25(2x)
16(x2 + y2) = 25
x2 + y2 = 25/16
y2 = (25/16) - x2 --- (2)
Substitute the value of y2 in (1)
8(x2 + (25/16) - x2)2 = 25[x2 - ((25/16) - x2)]
8(25/16)2 = 25(2x2 - (25/16))
On simplification,
5000/256 = 50x2 - 625/16
50x2 = (5000+10000)/256
50x2 = 15000/256
x2 = 300/256
Taking square root
x = ±(√300)/16
To find y² substitute the value of x2 in (2)
y2 = (25/16) - (300/256)
y2 = (400-300)/256
y2 = 100/256
y2 = 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(x2 + y2)2 = 25(x2 - y2)
Summary:
The points on the lemniscate 8(x2 + y2)2 = 25(x2 - y2) where the tangent is horizontal are ((√300, 5/8) ((√300, -5/8) (-(√300, 5/8) and (-(√300, -5/8).
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