JEE Main 2026 Mathematics question
Let a focus of the ellipse E : x²/a²+y²/b²=1 be S(4,0) and its eccentricity be 4/5. If the point P(3,α) lies on E and O is the origin, then the area of △POS is equal to:
Answer: (C)
Solution
c=4=ae gives a=5, b²=25(1−16/25)=9. At x=3: α²/9 = 1−9/25 = 16/25, so α=12/5. Area = ½×OS×|α| = ½×4×(12/5) = 24/5.