College algebra! Please help!

Question

A rectangular playing field lies in the interior of an elliptical track that is 30 yards wide and 240 yards long. What is the width of the rectangular playing field if the width is located 15 yards from either vertex?

Answer 1

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