The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance between the foci is (axes are coordinate axes)

We have     2b2a=8 and     2b=12(2ae)∴    2aae22=8 or     ae2=16----1 Also,     2×b2a=8 or     b2=4a or     a2e2−1=4a or     ae2−a=4----2From (1) and (2), we have 16−16e2=4or  16e2=12or   e=23