JEE Main Mathematics Question (2026) — Solution

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, the length is 4, and the height is the absolute value of the y-coordinate of P, which is 12 5 . Area = 1 2 4 12 5 = 24 5 Answer: 24/5