Ellipse Sector Area

Let the intersection of the line originating at (0,-1) and the positive part of the ellipse in the first quadrant be (x 0,y 0). Let P be the intersection of the x axis be denoted by P, and for convenience, we shall also call the distance from the origin to point P as P. Let F = point(x 0,y 0), O = origin, D = (0,-1), and E = (0,1), and finally G = (x 0,0). Then the area EDF will be 1/4 of the area of the ellipse , which is 5*pi/4. The equation of the ellipse is x^2/25 + y^2/1 = 1.The area EDF = area of OEFG - area of triangle PFG + area of triangle OFD.These areas are respectfully given as integral from 0 to x 0 of y(dx) - (1/2)(y 0)(X 0 - p) + (1/2)p..solving for y, y =sqrt(25-x^2)/5.P can be found in terms of x 0,y 0 from x 0 -P =sqrt((y 0)^2 + (x 0 - P)^2) - (y 0)^2, which yields P = (x 0)/((y 0) + 1). Using standard integration techniques(substitution), we find the necessary equation which must be solved for x 0. It Is (5/2)*{arcsin(x 0)/5) + ((x 0)/25)*sqrt(25 - (x 0)^2))} + ((x 0)/2)*[1 -sqrt(25 - (x 0)^2)/5] = 5*pi/4. The equation is solved numerically, giving a root close to x 0 = 3.69. Then y 0 = .6748, and P =2.203. Finally tan(theta/2) = P/1, and Theta rounded is 131. Ed Gray