0

A carpenter wishes to construct an elliptical table top from a sheet 4ft by 8ft plywood to make a poker table for him and his budies. He will trace out the ellipse using the "thumbtack and string" method. What length of string should be used and how far apart should the tacks be located, if the ellipse is to be the largest possible that can be cut out of the plywood sheet?

I encountered this problem at my school and I'm having some trouble with it. Alright, so what I know already. The largest possible ellipse will have a minor axis of 4 feet and a major axis of 8 feet, so the a and b values of the ellipse equation will be 2 and 4, respectively.

The equation then will be: x^2/16 + y^2/4 = 1

The vertices will be: (4,0), (-4,0), (0,2), (0,-2)

The combined distance from the foci to any point on the ellipse is the same, so the foci's coordinates can be (-f,0) and (f,0).

I am not sure what to do from this point.

conics123
  • 11
  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Please consider rewriting your post. – AlexR Mar 13 '15 at 00:03
  • Why would somebody downgrade this problem (though you really should tell us what you have tried thus far)? – Mark Fischler Mar 13 '15 at 00:03
  • Thank you for the input. I've written what I have tried thus far. – conics123 Mar 13 '15 at 00:07
  • Hint: The thumbtack-and-string construction produces an ellipse with the major axis equal to the length of the string. – hmakholm left over Monica Mar 13 '15 at 00:20

2 Answers2

1

Your next step, to fint $f$, is to note that the distance to the midpoint of one of the short sides of the table from one focus, and then back to the other (this goes right along the long axis), is $$ (f+4) + (4-f) = 8$$

But if instead you go to the midpoint of a long side and then to the other, you get a total distance of $$ \sqrt{f^2+2^2} + \sqrt{f^2+2^2} $$

Set that equal to $8$ and solve for $f$.

Mark Fischler
  • 41,743
1

Consider the following diagram:

enter image description here

From the property of the ellipse, $F_1B + BF_2 = F_1C+CF_2$. But $CF_2 = AF_1$, so that $F_1B + BF_2 = AC = 8$.

By the Pythagorean Theorem, $F_1B = BF_2 = \sqrt{F_1O^2+4}$. Then \begin{equation*} 2\sqrt{F_1O^2+4} = 8\ \Rightarrow\ 4(F_1O^2+4) = 64\ \Rightarrow\ F_1O^2 = 12 \ \Rightarrow F_1O = \sqrt{12}. \end{equation*} Then the foci are at $(\pm\sqrt{12},0)$.

rogerl
  • 22,399