An astronomer told his granddaughter how she could draw an ellipse by sticking two pins in a sheet of paper both equidistant from and in line with the central point, placing a loop of string around them, and then by putting a pencil in the loop and keeping it tight, go right round. The pencil would thus draw the ellipse, each pin being at a focus.

He then asked her to draw the biggest ellipse she could on a sheet of paper 50 centimetres long and 30 cm wide, and then to find the area of the rectangle she could inscribe in her ellipse whose longest side was equal in length to that of the shorter side of the original sheet.

How far apart must she place the pins? How long was the string? What was the area of the inscribed rectangle?

Note: Adapted from Enigma # 1733, which appeared in New Scientist on 23 January, 2013.

The standard equation for an ellipse centered about the origin is x^2/a^2 + y^2/b^2 =1.  a and b are the half lengths of the major and minor axes, and the distance between the foci is 2c, where c^2=a^2-b^2 (WOLG  a>b).  The area of the ellipse is pi*a*b.<o:p></o:p>

Given the paper size, 50x30, maximizing the area means maximizing a*b, but they ae independent, so a=25 and b=15, resulting in c=20. So:<o:p></o:p>

1)      The pins must be 40cm part<o:p></o:p>

2)      Since the string must stretch from one pin, along the major axis, past the second pin to edge of the paper and back to the original pin it must be 2*(40 + 5) = 90 cm long<o:p></o:p>

3)      No answer yet<o:p></o:p>

Edited on October 10, 2023, 12:14 pm

Posted by Kenny M on 2023-10-10 12:12:38