Let a focus of the ellipse E : x2a2+ y2b2=1 be S(4, 0) and its eccentricity be 45. If the point P(3, ) lies on E and O is the origin, then the area of POS is equal to:
- 125
- 145
- 245
- 485
Answer
(3) 245
Solution
[figure]
Focus S(ae, 0)=(4, 0), e= 45 a=5
b2=a2 (1-e2 )=25 925=9
P(3, ) on E: 925+ 29=1 2= 14425 | |= 125
Area of POS= 12 OS | |= 12 4 125= 245
Answer: (3)