If the eccentricity e of the hyperbola x2a2- y2b2=1, passing through (6, 4 3 ), satisfies 15 (e2+1 )=34e, then the length of the latus rectum of the hyperbola x2b2- y22 (a2+1 )=1 is :
- 10
- 20
- 25
- 30
Answer
(1) 10
Solution
15e2 - 34e + 15 = 0 (3e - 5)(5e - 3) = 0, so e = 53 or 35.
for a hyperbola e > 1, e = 53, and so
b2 = a2 (e2 - 1 ) = a2 ( 259 - 1 ) = 16a29
(6, 4 3 ) lies on the hyperbola, so 36a2 - 48b2 = 1
36a2 - 48 916a2 = 1 36 - 27a2 = 1
a2 = 9, b2 = 16
The second hyperbola is then x216 - y22(9 + 1) = 1, i.e. x216 - y220 = 1, with A2 = 16, B2 = 20.
Length of latus rectum = 2B2A = 2 204 = 10
Answer: (1)