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

Options

A. `7`
B. `8`
C. `10`
D. `9`

Explanation

For the hyperbola `(x^(2))/(144)-(y^(2))/(81)=(1)/(25)`, we have
`a^(2)=(144)/(25)`, `b^(2)=(81)/(25)`
`:.e^(2)=1+(b^(2))/(a^(2))=1+(81)/(144)=(15^(2))/(12^(2))impliese=(15)/(12)=(5)/(4)`
So, the coordinates of foci are `(+-ae,0)` i.e. `(+-3,0)`
For the ellipse `(x^(2))/(16)+(y^(2))/(b^(2))=1`, we have `a^(2)=16`
`:.e^(2)=1-(b^(2))/(16)=(16-b^(2))/(16)impliese=(sqrt(16-b^(2)))/(4)`
So, the coordinates of foci are `(+-sqrt(16-b^(2)),0)`
It is given that the ellipse and hyperbola have the same foci.
`:.sqrt(16-b^(2))=3impliesb^(2)=7`.