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# What is the x and y value of the answer to this problem?

Find the point (in the first quadrant) on the lemniscate 2(x^2+y^2)^2=25(x^2−y^2)

where the tangent is horizontal.

### 2 Answers

- 6 months agoFavorite Answer
Horizontal tangent means that dy/dx = 0

2 * (x^2 + y^2)^2 = 25 * (x^2 - y^2)

2 * 2 * (x^2 + y^2) * (2x * dx + 2y * dy) = 25 * (2x * dx - 2y * dy)

4 * (x^2 + y^2) * 2 * (x * dx + y * dy) = 25 * 2 * (x * dx - y * dy)

4 * (x^2 + y^2) * (x + y * (dy/dx)) = 25 * (x - y * (dy/dx))

dy/dx = 0

4 * (x^2 + y^2) * (x + y * 0) = 25 * (x - y * 0)

4 * (x^2 + y^2) * x = 25 * x

x = 0 is a possible solution to the set, so we can remove it

4 * (x^2 + y^2) = 25

4x^2 + 4y^2 = 25

4y^2 = 25 - 4x^2

2y = +/- sqrt(25 - 4x^2)

y = (+/- 1/2) * sqrt(25 - 4x^2)

We want the first quadrant, so y > 0

y = (1/2) * sqrt(25 - 4x^2)

y^2 = (1/4) * (25 - 4x^2)

2 * (x^2 + y^2)^2 = 25 * (x^2 - y^2)

2 * (x^2 + (1/4) * (25 - 4x^2))^2 = 25 * (x^2 - (1/4) * (25 - 4x^2))

2 * (x^2 + 25/4 - x^2)^2 = 25 * (x^2 - 25/4 + x^2)

2 * (25/4)^2 = 25 * (2x^2 - 25/4)

2 * 625/16 = 25 * (2x^2 - 25/4)

625/8 = 25 * (2x^2 - 25/4)

25/8 = 2x^2 - 25/4

25/4 + 25/8 = 2x^2

50/8 + 25/8 = 2x^2

75/8 = 2x^2

75/16 = x^2

x = +/- 5 * sqrt(3) / 4

x > 0

x = 5 * sqrt(3) / 4

y = (1/2) * sqrt(25 - 4x^2)

y = (1/2) * sqrt(25 - 4 * 75/16)

y = (1/2) * sqrt(25 - 75/4)

y = (1/2) * sqrt(100/4 - 75/4)

y = (1/2) * sqrt(25/4)

y = (1/2) * (5/2)

y = 5/4

(5 * sqrt(3) / 4 , 5/4)

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