Let the eccentricity e of a hyperbola satisfy the equation 6e2-11e+3=0. If the foci of the hyperbola are (3, 5) and (3, -4), then the length of its latus rectum is :
- 113
- 173
- 152
- 172
Answer
(3) 152
Solution
[figure]
6e2-11e+3=0 (2e-3)(3e-1)=0 e= 32 or 13
e>1 for a hyperbola e= 32
Foci S1(3, 5) and S2(3, -4) lie on x=3 (transverse axis vertical)
2ae=S1S2=9 a= 92e=3
b2=a2 (e2-1 )=9 ( 94-1 )= 454
Length of latus rectum = 2b2a= 23 454= 152
Answer: (3)