Posted by **Anonymous** on Tuesday, December 13, 2011 at 9:43pm.

A carpenter is cutting a 3ft by 4ft elliptical sign from a 3ft by 4ft piece of plywood. The ellipse will be drawn using a string attached to the board at the foci of the ellipse.

A)How far from the ends of the board should the string be attached?

B)How long should the string be?

- College Algebra -
**drwls**, Wednesday, December 14, 2011 at 1:52am
The semimajor axis is a = 2 ft.

The semiminor axis is b = 1.5 ft.

The equation of the ellipse is

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

I have assumed (0,0) is the center of the rectangle.

The foci are at y = 0 and x = +/- c, where

c = sqrt[a^2 - b^2] = sqrt1.75

= 1.323 ft

The distance of the foci from the closest end of the board is 4 - 1.323 = 2.677 ft

String length = 2c = 2*s

- analytic geometry -
**drwls**, Wednesday, December 14, 2011 at 1:55am
Ignore the final '=2s '

The answer got sent before I could finish.

2c = 2.646 ft is the string length

- College Algebra -
**Anonymous**, Wednesday, December 14, 2011 at 11:52am
the sum of one half a number, one third a number and one twelfth of a number is 33. find the original number

- College Algebra -
**ajchat**, Wednesday, May 4, 2016 at 7:39pm
Slight mistake in the first answer...The total distance the foci are apart from each other is 2.6457 ft. BUT the distance of the foci from the closest end of the board is 2 - 1.32288 = .67712 ft. Since c^2= a^2 - b^2, and 2a=4, and a=2, therefore a-c is equal to 2-1.32288 which is only .67712 ft. Also, string length is equal to 2c + 2(a-c)=2c + 2a - 2c which is only equal to 2a. So, the string length is 4 ft.

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