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)`

Options

A. the equation of hyperbola is `(x^(2))/(9)-(y^(2))/(16)=1`
B. the equation of the hyperbola is `(x^(2))/(9)-(y^(2))/(25)=1`
C. the vertex of the hyperbola is (5, 0)
D. the vertex of the hyperbola is `(5sqrt3, 0)`

Answer

A, C

Explanation

for given ellipse
`(x^(2))/(25)+(y^(2))/(16)=1`
we have
`e=sqrt(1-(16)/(25))=(3)/(5)`
Hence, the eccentricity of the hyperbola is `5//3`.
Let the hyperbola be
`(x^(2))/(A^(2))-(y^(2))/(B^(2))=1`
Then `B^(2)=A^(2)(e^(2)-1)=A^(2)((25)/(9)-1)=(16)/(9)A^(2)`
Therefore, the equation of the hyperbola is
`(x^(2))/(A^(2))-(9y^(2))/(16A^(2))=1`
As it passes through (3, 0), we get `A^(2)=9 and B^(2)=16.`
The equation is
`(X^(2))/(9)-(y^(2))/(16)=1`
The foci of the hyperbola are `(pmae, 0)-=(pm 5, 0).`
The vertex of the hyperbola is (3, 0).