Let a focus of the ellipse E: x^2 a^2 + y^2 b^2 = 1 be S(4, 0) and its eccentricity be 4 5 . If the point P(3, ) lies on E and O is the origin, then the area of POS is equal to:

Step-by-step solution

Given the focus of the ellipse S(4, 0) , we have ae = 4 . Since the eccentricity e = 4 5 , we get a ( 4 5 ) = 4 a = 5 . Using the relation b^2 = a^2(1 - e^2) , we find: b^2 = 25 (1 - 16 25 ) = 9 The equation of the ellipse is x^2 25 + y^2 9 = 1 . Since the point P(3, ) lies on the ellipse, substituting x = 3 gives: 9 25 + ^2 9 = 1 ^2 9 = 1 - 9 25 = 16 25 ^2 = 144 25 | | = 12 5 The coordinates of the vertices of POS are O(0, 0) , S(4, 0) , and P (3, 12 5 ) . The area of POS is 1 2 base height . Taking OS as the base