In a hyperbola the distance between the foci is 2 and the distance between directrices is 1 Its eccentricity is
√3
32
√2
52
ss1=2⇒2ae=2
ZZ1=1⇒2ae = 1
→ e2=2 ⇒ e = √2
In a hyperbola the distance between the foci is three times the distance between the directories then its eccentricity is