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标题: Area of an ellipse [打印本页]

作者: yma16 时间: 2005-4-30 00:17
标题: Area of an ellipse

Find the area of the smallest ellipse that passes (1,0), (1,1), (-1,0) and (-1,1).

作者: 野菜花 时间: 2005-4-30 01:25
标题: 回复：Area of an ellipse

pi ?

作者: yma16 时间: 2005-4-30 07:50
标题: Sorry, I don't know. Pls show it.

Sorry, I don't know. Pls show it.

作者: yma16 时间: 2005-5-1 00:54
标题: right. Could you try another?

I verified it and it is pi. There is a similar question which asking to find the area of min ellipse to enclose 2 circles with r=.5 and centers at (.5,0) and (-.5,0). I'll try it when I have time.

Thank you very much for contributing so many questions and answers, and articles in this and other boards.

作者: 野菜花 时间: 2005-5-1 07:56
标题: 回复：right. Could you try another?

Sorry, since I went to church for whole day, I did not have chance to show the calculation for your 1st problem.

Thanks for your encouragement!

The 2nd one:

Let the equation of the ellipse be

x^2/a^2+y^2/b^2=1

Consider the intersection between the ellipse and the circle

(x-0.5)^2 + y^2 =0.5^2 (i.e. x^2 - x + y^2=0 )

Substituting y^2=x-x^2 into the equation of the ellipse, we got

x^2/a^2+(x-x^2)/b^2=1.

Simplifying it, we have

(b^2-a^2)*x^2 +a^2*x-a^2*b^2=0

As the quadratic equation has a unique solution for x, we have

a^4-4(a^2-b^2)*a^2*b^2=0,

so, a^2=4b^4/(4b^2-1)

Since the area of the ellipse is ab(pi), we may just find the min value of a^2*b^2.

a^2*b^2=4b^6/(4b^2-1)

use derivative, we found

ab reaches the minimum when b=(3/8)^0.5, a=3/8^0.5

The min of the area is

3/8*sqrt3*pi

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