Vertices `(pm2,0)`, foci `(pm3,0)`
`rArr` The transverse axis of hyperbola is along x-axis.
`:.a=2" "and" "ae=3`
Now, `b^(2)=a^(2)(e^(2)-1)`
`=a^(2)e^(2)-a^(2)`
`=3^(2)-2^(2)=5`
and equation of hyperbola
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`
`rArr(x^(2))/(2^(2))-(y^(2))/(5)=1rArr(x^(2))/(4)-(y^(2))/(5)=1`