InterviewSolution
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InterviewSolution
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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 themajor and minor axes of the ellipse and if the product of eccentricities ofhyperbola and ellipse is 1 thenthe equation ofhyperbola 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. focusof hyperbola is `(5sqrt(3),0)`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)` |
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Answer» Correct Answer - A::C 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). |
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