Let the product of the focal distances of the point ( 3, 12 ) on the ellipse x2a2+ y2b2=1, (a>b), be 74. Then the absolute difference of the eccentricities of two such ellipses is
- 1-2 2 3
- 1- 3 2
- 3-2 22 3
- 3-2 23 2
Answer
(3) 3-2 22 3
Solution
Focal distances of P(x1,y1) are a ex1, so their product is a2-e2x12.
[figure]
a2-3e2= 74 4a2=7+12e2; P on the ellipse with b2=a2(1-e2):
12(1-e2)+1=4a2(1-e2) 12e4-17e2+6=0
e2= 23, 34, so |e1-e2|= 32- 23= 3-2 22 3
Answer: (3)