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

Options

A. 1
B. 5
C. 7
D. 9

Answer

C

Explanation

Given equation of ellipse is `(x^(2))/(16)+(y^(2))/(b^(2))=1`
Here, `a^(2)=16 Rightarrow a=4`
`therefore c=sqrt(1-(b^(2))/(16))=(sqrt(16-b^(2)))/(4)`
`therefore` Foci of ellipse are `(pm ae, 0)ie, (pm sqrt(16-b^(2)),0)`
Also, given equation of hyperbola is `(x^(2))/(144)-(y^(2))/(81)=(1)/(25)`
`"Here", a^(2)=((12)/(5))^(2),b^(2)=((9)/(5))^(2)`
`therefore esqrt(1+(b^(2))/(a^(2)))=sqrt(1+(81)/(144))=(5)/(4)`
`therefore ` Foci of the hyperbola are `(pm ae,0)ie, (pm, 3,0)`
According to the given condition
foci of ellipse=foci of hyperbola
`therefore sqrt(16-b^(2))=3`
`Rightarrow b^(2)=7`