If `a` hyperbola passes through the foci of the ellipse `(x^2)/(25)+(y^2)/(16)=1` . Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse and if the product of eccentricities of hyperbola and ellipse is 1 then the equation of hyperbola is `(x^2)/9-(y^2)/(16)=1` b. the equation of hyperbola is `(x^2)/9-(y^2)/(25)=1` c. focus of hyperbola is (5, 0) d. focus of hyperbola is `(5sqrt(3),0)`

Explanation

Here, equation of ellipse is,

`x^2/25+y^2/16 = 1`

So, if e is the essentricity of ellipse, then,

`e^2 = 1-b^2/a^2 = 1-16/25 = 9/25`

`e = 3/5`

We know, foci of ellipse is (ae,0). So, focus of given ellipse will be `(3,0)`.

Now, let equation of hyperbola is,

`x^2/a^2-y^2/b^2 = 1`

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