Make Poker Table

Full question:
A carpenter wants to build an elliptical table with a sheet of 4 'x 8' plywood to make a poker table for him and his budies. He draw the ellipse using the bug "and the string" method. What is the length of the string must be used and at what distance should be the nails, if the ellipse is to be as much as possible that can be cut into the plywood?

The greatest possible ellipse has a semi-minor axis of 2 feet and a semi-major axis of 4 feet. If we center the board on the origin of the Cartesian plane, one can infer the location of homes and, consequently, the length of the string, you must:

x ^ 2 / 16 + y ^ 2 / 4 = 1

Summits:

(-4, 0), (4, 0)
(0, -2), (0, 2)

Now we know that the combined distance of homes at any point of the ellipse is the same. If we fix the homes following coordinates: (-f, 0) and (f, 0), we can determine this issue:

First, find the combined distance along the major axis:

D (-4,-f) = m
D (-4, f) = 2F + m

m + m + 2f = 2m + 2f


Now we divide by 2 and set (f + m) length of the hypotenuse of a right triangle that has vertices on the origin, on a summit of the ellipse, and attention

f ^ 2 + 2 ^ 2 = (f + m) ^ 2
f ^ 2 + 4 = f ^ 2 + m ^ 2 + 2mf
4 = m ^ 2 + 2mf
m ^ 2 + 2mf - 4 = 0

m = (-2F + / - sqrt (4f ^ 2 + 16)) / 2
m = (-2F + / - 2 * sqrt (f ^ 2 + 4)) / 2
m = f-+ / - sqrt (f ^ 2 + 4)

= f 4 - f + m = 4 - / + sqrt (f ^ 2 + 4)
0 = 4 + / - sqrt (f ^ 2 + 4)
-4 = Sqrt (f ^ 2 + 4)
16 = f ^ 2 + 4
12 = f ^ 2
2 * sqrt (3) = f

Thus, homes are located at (+ / - 2 * sqrt (3), 0)

The string length is 8 feet.