If the lengths of transverse and conjugate axis of the hypberbola are 4,2 then the distance Between the foci is
4√5
8√5
√5
2√5
2a = 4 ⇒ a = 1
2b = 2 ⇒ b = 1 e = √1+14=√52
ss1=2ae=2.2√52=2√5