If P be any point on the ellipse 16x^2 + 25y^2 = 400 with foci S and S' and area of PSS' is 9 square units, then the abscissa of point P is...........

Step-by-step solution

The equation of the ellipse is 16x^2 + 25y^2 = 400 , which can be written as x^2 25 + y^2 16 = 1 . Here, a^2 = 25 and b^2 = 16 . The eccentricity e = 1 - b^2 a^2 = 1 - 16 25 = 3 5 . The foci S and S' are at ( ae, 0) = ( 3, 0) . The distance between the foci is SS' = 6 . Let the coordinates of point P be (x, y) . The area of PSS' is given by 1 2 SS' |y| . Given that the area is 9 square units, we have: 1 2 6 |y| = 9 3|y| = 9 |y| = 3 Substituting y^2 = 9 into the equation of the ellipse: 16x^2 + 25(9) = 400 16x^2 + 2