Let `e_(1)` and `e_(2)` be the eccentricities of ellipse and hyperbola, respectively.
Since hyperbola passes through the foci of ellipse, we have
`b=ae_(1)" (1)"`
Also for ellipse, `b^(2)=a^(2)(1-e_(1)^(2))" (2)"`
and for hyperbola, `a^(2)=b^(2)(e_(2)^(2)-1)" (3)"`
From (1) and (2), we get
`2e_(1)^(2)=1 or e_(1)=(1)/(sqrt(2))`
So, from (1) and (3), we get
`2=e_(2)^(2)-1 or e_(2)=sqrt3`