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Given the ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1, find the maximum area of its inscribed rectangle
Let the vertex of the rectangle in the first quadrant be m (x, y), then the area of the rectangle is s = 4xy
Now find the maximum value of S = XY under the condition x ^ 2 / 9 + y ^ 2 / 4 = 1
From the inequality of arithmetic mean and geometric mean, it is obtained that
S=4xy=4*(x/3)*(y/2)*6
When the maximum area of the triangle with one point on the ellipse and two focal points of the ellipse as the vertex is 1, the minimum value of the major axis of the ellipse is ()
A. 22B. 2C. 2D. 22
We know that BC = 1. A2 = B2 + C2 = B2 + 1b2 ≥ 2, a ≥ 2. 2A ≥ 22
How to find the maximum perimeter of a gold ellipse inscribed rectangle
Let the major axis of the ellipse be 2a, the minor axis be 2B, and the side length of the rectangle be 2x, 2Y,
And x = ACOS θ, y = bsin θ, perimeter = 4x + 4Y = 4acos θ + 4bsin θ = 4 radical (a ^ 2 + B ^ 2) sin (θ + α)
The maximum perimeter of the inscribed rectangle of the golden ellipse is 4 radicals (a ^ 2 + B ^ 2) = 4A radicals (1 + 0.618...)
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