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 e= 53 or 35
For a hyperbola e>1 e= 53
b2=a2 (e2-1 )=a2 ( 259-1 )= 16a29
(6, 4 3 ) lies on the hyperbola : 36a2- 48b2=1
36a2- 48 916a2=1 36-27a2=1
a2=9, b2=16
Second hyperbola : x216- y22(9+1)=1, i.e. x216- y220=1
Here A2=16, B2=20; length of latus rectum = 2B2A= 2 204=10
Answer: (1)