standard equation for an ellipse is

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

vertices are (-a,0) and (a,0)

a=240/2 = 120

equation for the elliptical track is

x^/120^2 + y^2/15^2 = 1

since the rectangular field is 15 from the vertices,

it's location is (120-15,0) = (105,0)

It's traditional to graph standard ellipses with center at the origin

now plug in 105 into the ellipse equation to solve for y

105^2/120^2 + y^2/15^2 = 1

y^2/15^2 = 1 - 105^2/120^2 = (120^2 - 105^2)/120^2 = 13475/120^2

y^2 = 15^2 times 3475/120^2

y = + or - 15/120 times square root of 3475

= almost + or - (1/8)(29) = 29/8 = 3.625

double y to get the width of the rectangle 2(29/8) = 58/8 = 7.25 yards

the rectangle is 7 1/4 yards by 210 yards

but you said 15 yards from "either" vertex. The above is correct for either horizontal vertex. But if we use the vertical vertices, that should give a different answer. 15 yards from the vertices (0,15) and (0,-15) creates a different side, a pinpoint for the "width" of zero. With a length of 240 yards. A 0 by 240 rectangle, which isn't really a rectangle at all, just a line. or two lines in one

Sierra H.

thank you SO much!!04/18/20