A hyperbola passes through a focus of the ellipse `x^2/169 + y^2/25=1.` Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse.The product of eccentricities is 1. Then the equation of the hyperbola is

Options

A. `(x^(2))/(144)+(y^(2))/(9)=1`
B. `(x^(2))/(169)-(y^(2))/(25)=1`
C. `(x^(2))/(144)-(y^(2))/(25)=1`
D. `(x^(2))/(25)-(y^(2))/(9)=1`

Answer

C

Explanation

Let the equation of hyperbola be
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` . . (i)
given equation of ellipse is
`(x^(2))/((13)^(2))+(y^(2))/((5)^(2))=1`
here, `a=13,b=5`
`thereforee=sqrt(1-(b^(2))/(a^(2)))`
`=sqrt(1-(25)/(169))=sqrt((144)/(169))=(12)/(13)`
`therefore` Focus (`+-` aw, 0)=`(+-13xx(12)/(13),0)`
`=(+-12,0)`
Since, Eq. (i) passes through `(+-12,0)`
`therefore(144)/(a^(2))-(0)/(b^(2))=1`
`impliesa^(2)=144impliesa=+12`
Now, eccentricity of hyperbola
`e'=sqrt(1+(b^(2))/(b^(2)))=sqrt(1+(b^(2))/(144))`
According to the equation, `e e'=1`
`(12)/(13)xxsqrt(1+(b^(2))/(144))=1`
`implies sqrt(1+(b^(2))/(144))=(13)/(12)`
`implies1+(b^(2))/(144)=(169)/(144)`
`implies(b^(2))/(144)=(169)/(144)-1`
`implies(b^(2))/(144)=(25)/(144)impliesb^(2)=25`
`therefore` Equation of hyperbola is
`(x^(2))/(144)-(y^(2))/(25)=1`.