Vertices `-=(pm7,0)ande=(4)/(3)` Mbrgt Vertices lie on x-axis.
`:.` The transverse axis of hyperbola is along the x-axis.
Therefore, `a=7rArra^(2)=49`
and `b^(2)=a^(2)(e^(2)-1)`
`=7^(2)((16)/(9)-1)=(343)/(9)`
Equation of hyperbola
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`
`rArr(x^(2))/(49)-(y^(2))/(343//9)=1rArr(x^(2))/(49)-(9y^(2))/(343)=1`