Yahoo Answers: Answers and Comments for What is the x and y value of the answer to this problem? [Mathematics]
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From Anonymous
enUS
Mon, 15 Jul 2019 23:44:52 +0000
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Yahoo Answers: Answers and Comments for What is the x and y value of the answer to this problem? [Mathematics]
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From alex: Rule
Horizontal tangent <> dy/dx = 0
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Tue, 16 Jul 2019 00:44:01 +0000
Rule
Horizontal tangent <> dy/dx = 0

From Captain Matticus, LandPiratesInc: Horizontal tangent means that dy/dx = 0
2 * (...
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Tue, 16 Jul 2019 00:08:37 +0000
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)