If the foci of the ellipse `(x^2)/(16)+(y^2)/(b^2)=1` and the hyperbola `(x^2)/(144)-(y^2)/(81)=1/(25)` coincide, then find the value of ` b^2`

Explanation

Given hyperbola is
`(x^(2))/(144)-(y^(2))/(81)=(1)/(25)`
`"or "(x^(2))/(144//25)-(y^(2))/(81//25)=1`
So, foci of hyperbola are `(pmsqrt((144)/(25)+(81)/(25)),0)-=(pm3,0)`
These are foci of ellipse `(x^(2))/(16)+(y^(2))/(b^(2))=1` also.
`therefore" "3=sqrt(16-b^(2))`
`therefore" "b^(2)=7`